用消元法解方程；10X+Y+2Z=1,25X-3Y-Z=4,5X+5Y+5Z=2, 用消元法解方程；10X+Y+2Z=1,25
用消元法解方程；10X+Y+2Z=1,25X-3Y-Z=4,5X+5Y+5Z=2
许叶学生 用消元法解方程；10X+Y+2Z=1,25X-3Y-Z=4,5X+5Y+5Z=2
10X+Y+2Z=1
(1)25X-3Y-Z=4
(2)5X+5Y+5Z=2
(3)(1)+(2)*210X+Y+50X-6Y=1+860X-5Y=9
(4)(2)*5+(3)125X-15Y+5X+5Y=20+2130X-10Y=22
(5)(5)-(4)*2130X-120X=22-福骇弟较郗记甸席鼎芦1810X=4X=2/5Y=(60X-9)/5=3Z=25X-3Y-4=-3
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求过点A(0,6)且与圆C:x2+y2+10x+10y=0切于原点的圆的方程.
活动:学生思考交流,回顾圆的方程的求法,教师引导学生注意题目的条件,灵活处理,如图3.所求圆经过原点和A(0,6),且圆心应在已知圆的圆心与原点的连线上.根据这三个条件可确定圆的方程.图3解:将圆C化为标准方程,得(x+5)2+(y+5)2=50,则圆心为C(-5,-5),半径为.所以经过此圆心和原点的直线方程为x-y=0.设所求圆的方程为(x-a)2+(y-b)2=r2.由题意,知O(0,0),A(0,6)在此圆上,且圆心M(a,b)在直线x-y=0上,则有解得于是所求圆的方程是(x-3)2+(y-3)2=18.点评:求圆的方程,一般可从圆的标准方程和一般方程入手,至于选择哪一种方程形式更恰当,要根据题目的条件而定,总之要让所选择的方程形式使解题过程简单.
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THE BINOMIAL THEOREM
THE BINOMIAL THEOREM shows how to calculate a power of a binomial -- (a + b)n -- without actually multiplying out.
For example, if we actually multiplied out the 4th power of (a + b) --
(a + b)4 = (a + b)(a + b)(a + b)(a + b)
-- then on collecting like terms we would find:
b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 . & .&
Notice: &The literal factors are all the combinations of a&and&b where the sum of the exponents is 4: &a4, &a3b, &a2b2, &ab3, &b4.
of each term is 4.
The first term is actually a4b0, which is a4& 1.
to &expand& (a + b)5, we would anticipate the following terms, in which the sum of all the exponents is 5:
(a + b)5 = &?&a5 + &?&a4b + &?&a3b2 + &?&a2b3 + &?&ab4 + &?&b5
The question is, What are the coefficients?
They are called the binomial coefficients. &In the expansion of (a + b)4, the binomial coefficients are &1
&1; & above.& Note the symmetry: &The coefficients from left to right are the same right to left.
The answer to the question, "What are the binomial coefficients?" is called the binomial theorem. &It shows how to calculate the coefficients in the expansion of (a + b)n.
The symbol for a binomial coefficient is . &The upper index n is the expon the lower index k indicates which term, starting with k = 0.
The lower index k is the .
= 5, each term in the expansion of &(a + b)5 &will look like this:
k will successively take on the values 0 through 5.
(a + b)5 =
a5 &+ & a4b &+ & a3b2 &+ & a2b3 &+ & ab4 &+ & b5
Again, each lower index
is the exponent of b. &The first term has k&=&0 because in the first term, b appears as b0, which is 1.
Now, what are these binomial coefficients,
The theorem states that the binomial coefficients are none other than the , nCk .
& (a + b)5
5C0 a5 + 5C1 a4b + 5C2 a3b2 + 5C3 a2b3 + 5C4 ab4 + 5C5 b5
1a5 + a4b + a3b2 + a2b3 + ab4 + b5
a5 &+& 5a4b &+& 10a3b2 &+& 10a2b3 &+& 5ab4 &+& b5
The binomial coefficients here are &1& 5& 10& 10& 5& 1.& Note the symmetry.
The coefficient of the first term is always 1, and the coefficient of the second term is the same as the exponent of (a + b), which here is 5.
Note also that, when the
combinatorial number is written as factors, each coefficient contains the previous one. &The coefficient of a3b2 is the previous coefficient multiplied by 4 and divided by 2.
That is, it is multiplied by the exponent of the previous a, and divided by the present exponent of b.
The coefficient of a2b3 is the previous coefficient multiplied by 3 and divided by 3. &And so on. &We will see this in .
for the combinatorial numbers, here is the binomial theorem:
What follows the summation sign is the general term. &Each term in the sum will look like that -- the first term having k = 0; then k = 1, k&=&2, and so on, up to k = n.
Notice that the sum of the exponents &(n & k) + k, always equals n.
Example 1.
a) &The term &a8b4 &occurs in the expansion of what binomial?
Answer.&&&(a + b)12. &The sum of 8 + 4 is 12.
b) &In that expansion, what number is the coefficient of a8b4?
&Answer.&&&It is the combinatorial number,
12&&11&&10&&9&&&1&&2&&3&&4
Note again: The
, in this case 4, is the exponent of b.
This same number is also the coefficient of &a4b8, &since &12C8 = 12C4.
Example 2.&&& (a & b)5.
&Solution.&&&We found the
to be &1 &5& 10& 10& 5& 1. &The difference with (a & b) is that the signs of the terms will alternate:
(a & b)5 = a5 & 5a4b + 10a3b2 & 10a2b3 + 5ab4 & b5.
For, &a & b = a + (&b), &therefore each term will have the form
a5 & k(&b)k.
When k is even, (&b)k will be positive. &But when k is odd, (&b)k will be negative. &
Each odd power of b will have a minus sign.
Example 3.&&&In the expansion of &(x & y)15, calculate the coefficients of x3y12&and& x2y13.
&Solution.&&&The coefficient of& x3y12 &is positive
because the exponent of
y is even. &That coefficient is 15C12. &But 15C12 = 15C3, and so we have
15&&14&&13 &&&&1&&2&&3
The coefficient of &x2y13, on the other hand, is negative because the exponent of y is odd. &The coefficient is && 15C13
&=& & 15C2. &We have
15&&14&&1&&2
Example 4.&&& the first four terms of (a + b)n. &Do not use factorials.
&&&(a + b)n
+ &nC1 an & 1b&
+ &nC2 an & 2b2&
+ &nC3 an & 3b3
+ &nan & 1b&
+ an & 2b2&
+ &an & 3b3.
Notice: &Each coefficient is a factor of the next coefficient. &The coefficient n is a factor of
That in turn is a factor of
To construct the next coefficient, then, multiply the present coefficient by the exponent of a in that term --
-- namely n & 3:
n(n & 1)(n & 2)(n & 3)&&&&&&&&&1&&2&&3&&4
And divide it by 1 more than the exponent of b.
That is the coefficient of an & 4b4.
Example 5.&&& the binomial theorem to expand &(a + b)8.
&Solution.&&&The expansion will begin :
(a + b)8 = a8 + 8a7b
The first coefficient is always 1. &The second is always the exponent of the expansion, in this case 8.
The next coefficient can be constructed as described . &It will be the present coefficient, 8 --
(a + b)8 = a8 + 8a7b + 28a6b2
-- times the exponent of a in that term , 7, divided by 1 more then the exponent of b. &It will be 8&&7/2 = 28.
The next coefficient --
(a + b)8 = a8
-- is 28&&6, divided by 3:
28&&6 / 3 = 28&&2 = 56.
The next --
(a + b)8 = a8
-- is 56&&5, divided by 4:
56&&5 / 4 = 14&&5 = 70.
We have now come to the point of symmetry.& For, the coefficient of a3b5 is equal to the coefficient of a5b3, which is 56. &And so on for the remaining coefficients.
Here is the complete expansion:
a8&+&8a7b&+& 28a6b2&+&56a5b3&+&70a4b4&+& 56a3b5&+
28a2b6&+&8ab7&+&b8.
Example 6.&&&Write the 5th term in the expansion of (a + b)10.
&Solution.&&&In the 1st term, k = 0. &In the 2nd term, k = 1. &And so on.
The index k -- the exponent of b -- of each term is one less than the
of the term.
Thus in the 5th term, k = 4. &The exponent of b is 4. &The 5th term is
10&&9&&8&&71&&2&&3&&4
&a6b4&&=&&210 a6b4
This triangular array is called Pascal's Triangle. &Each row gives the combinatorial numbers, which are the binomial coefficients. &That is, the row &1 &2& 1& are the combinatorial numbers 2Ck, which are the coefficients of (a&+&b)2. &The next row,& 1 &3& 3& 1,& are the coefficients of (a + b)3; and so on. &
To construct the triangle, write 1, and below it write 1 &1. &Begin and end each successive row with 1. &To construct the intervening numbers, add the two numbers immediately above.
Thus to construct the third row, begin it with 1,
then add the two numbers immediately above: &1 + 1. &Write 2. &Finish the row with 1. &
To construct the next row, begin it with 1, and
the two numbers immediately above: 1 + 2. &Write 3. &Again, add the two numbers immediately above: &2 + 1 = 3. &Finish the row with 1.
Example 8.&&&Expand &(x & 1)6.
&Solution.&&&According to , the coefficients are
1& 6& 15& 20& 15& 6& 1.
In the binomial, x is &a&, and
&1 is &b&. &The
will alternate:
x6 & 6x5&&1
+ 15x4&&12
& 20x3&&13
+ 15x2&&14
x6 & 6x5 + 15x4 & 20x3 + 15x2 & 6x + 1
Example 9.&&&Expand &(x + 2)3.
&Solution.&&&The coefficients are &. & x is &a&, and 2 is &b&.
x3 + 3x2&&2 + 3x&&22 + 23
x3 + 6x2 + 12x + 8
Problem 1.&&& the expansion of (a + b)n, each term has the form
where k successively takes on the value 0, 1, 2, . . ., n.
is the symbol for the binomial coefficient.
The binomial theorem is the statement that & = ?
To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload").
The combinatorial number,&nCk.
Problem 2.&&&Use factorials to write the general term in the expansion of (a + b)n.
&&&&&&n!&&&&&&(n & k)! k!
Problem 3.&&&
a) & Calculate the coefficient of &a4b6 &in the expansion of (a + b)10.
10C6 = 10C4 = 210
b) & The coefficient of which other term is the same?&&
c) & In the expansion of &(a + b)n, the coefficient of &an & kbk& is the same
c) &&as the coefficient of which other term?&&
Problem 4.&&&Calculate the coefficient of
a) & x17y3 in the expansion of (x + y)20.&&
b) & x3y17 in the expansion of (x + y)20.&&
c) & x3y17 in the expansion of (x & y)20.&&
d) & x2y18 in the expansion of (x & y)20.&&
e) & x5y5 in the expansion of (x & y)10.&&
f) & x10 in the expansion of (x & 1)15.&&
Problem 5.&&& the first four terms of (x +
h)n. &Do not use factorials.
xn + nxn&1h +&
n(n & 1)&&1&&2
n(n & 1)(n & 2)&&&&&&1&&2&&3
Problem 6.&&&Compute the first four terms of each of the following.
a) & (a + b)15&&
15a14b + 105a13b2 + 455a12b3
b) & (x & 1)20&&
x20 & 20x19 + 190x18 & 1140x17
Problem 7.&&&Consider the expansion of (x + b)30.
a) & What is the exponent of b in the 1st term?&&
b) & What is the exponent of b in the 3rd term?&&
c) & In the 25th term?&&
d) & In the kth term?&&
e) & Write the fourth term, with its coefficient.&&
4,060x27b3
Problem 8.&&&Calculate each of the following.
a) & The third term of (a + b)11.&&
b) & The fifth term of (x & y)7.&&
c) & The tenth term of (x & 1)12.&&
& &d) & The fifteenth term of (1 +&
& &e) & The fourth term of (x &&
Problem 9.&&&Use Pascal's triangle to expand the following.
a) & (a + b)3&
= &a3 + 3a2b + 3ab2 + b3
= &a3 & 3a2b + 3ab2 & b3
= &x4 + 4x3y + 6x2y2 + 4xy3 + y4
= &x4 & 4x3y + 6x2y2 & 4xy3 + y4
= &x5 & 5x4 + 10x3 & 10x2 + 5x & 1
= &x5 + 10x4 + 40x3 + 80x2 + 80x + 32
= &8x3 & 12x2 + 6x & 1
= &1 & 7xy + 21x2y2 & 35x3y3 + 35x4y4 & 21x5y5 + 7x6y6 & x7y7
In the following Topic we will explain why the binomial coefficients
are the combinatorial numbers. That will constitute a proof of the binomial theorem.
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to keep TheMathPage online.Even $1 will help.
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Lawrence Spector
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& 32+x 2 37-x 0- 5 3 配方法解详细一点 (1)x2+10x+16=0 (2)x2-x-4/3=0 (3)3x2。
32+x 2 37-x 0- 5 3 配方法解详细一点 (1)x2+10x+16=0 (2)x2-x-4/3=0 (3)3x2。
收集整理：/ 时间：
配方法解详细一点 (1)x2+10x+16=0 (2)x2-x-4/3=0 (3)3x2。 【x+8】x【x+2】=0得--8.。--2
(1)x2+10x+16=0 x2+10x+25-25+16=0 (x+5)2=9两边开平方得:x+5=±3x1=-2 x2=-8(2)x2-x-4/3=0x2-x+1/4-1/4-3/4=0(x-1/2)2=1两边。已知函数f(x)=ax^3+1/2(sinθ)x^2-2x+c图像经过点(1,37/6),。(1)f(1)=37/6,即a+1/2sinθ-2+c=37/6 ①f(1)=0,即3a+sinθ-2=0 ②注意到f(-2)≤0,即12a-2sinθ-2≤0 ③由②③可得sinθ≥1,故sinθ=1解得a=1/3,c=22/3故f(x)=1/3x^3+1/2x^2-2x+22/3(2)首先无视m≥0这一条件。思路:命题等价于在一个长度为3的区间上,最大值与最小值的差小于37/6。f(x)在-2,1这两个极值点两侧的单调性为增,减,增,故先算中间的区间。注意到|f(-2)-f(-5)|=45/2,|f(4)-f(1)|=45/2,故m∈[-5,1]时符合题意(自己看看为什么,这里表述不清楚,不会再问我)接下来令g(m)=f(m+3)-f(m)(m不属于[-5,1]),解不等式|g(m)|≤45/2得-5≤m≤1,故当m不在[-5,1]时不合题意,综上m的取值范围为[0,1]
已知函数f(x)=ax3+(1/2)(sinθ)x2-2x+c图像经过点(1,37/6),且在[-2,1]内单调递减,在[1,+∞)上单调递增;(1)求f(x)解析;(2。已知函数f(x)的定义域是R,若f(x-2)是偶函数,f(x+7)也是偶。f(13)]【原题的[0,即(-5、f(3)。3,下面给出推证,如果以两条对称轴中间的一条为对称轴。和你的答案不一致,又1=7-6,7]有问题,我们已经看到y=f(x)的两条对称轴x=-2和x=72,则g(-x)=f(-x-2)由g(-x)=g(x),16]上我们只能保证有4个零点[f(1),由f(7+x)=f(7-x)。这个周期我们取[-2,y=f(x+7)的对称轴向右平移7个单位即得到y=f(x)的对称轴】由上面的讨论。所以y=f(x-2)的图像向右平移-2个单位(向左平移2个单位)即得到y=f(x)的图像,有4×112=448个零点。4;所以共有895个零点,从而y=f(x-2)的对称轴向左平移2个单位即得到y=f(x)的对称轴,并且也是对称轴、[-2?例如(1,18是它的一个正周期,因为不能确定[-2、设g(x)=f(x-2)!再看看f(1)=f(-2+3)。
f(x)=0的根为10n+1或10n+3的形式 ,解得0.且0≤n∴0≤10n+1≤2013:∵y=f(2-x)与y=f (7+x)都是偶函数.且0≤n-n≤0,10]上恰有两。圆C:(x-0.5)^2+(y-3)^2=37/4 -m .与直线:x+2y-3=0交于P,Q,。联立方程,得 (x-0.5)^2+(y-3)^2=37/4 -m ① x+2y-3=0 ② 将②代入①,化简得 5x^2+2x-27+4m=0 ③ ∵直线与圆交于两点 ∴③的△=4-4×5×(-27+4m)&=0 ∴m∈(-∞,68] 设P(x1,y1),Q(x2,y2) 根据韦达定理,得 x1x2=(-27+4m)/5 ,x1+x2=-2/5 y1y2=[(3-x1)/2]×[(3-x2)/2]=[9-3(x1+x2)+x1x2]/4 =[9-3×(-2/5)+(-27+4m)/5]/4=(m+6)/5 ∵向量OP乘以向量OQ=0 ∴x1x1+y1y2=0 即 (-27+4m)/5+(m+6)/5=0 解得 m=21/5 经检验,m=21/5时,③的△&0,满足题意。 所以,常数m的值为21/5。求y=[(2^x)-4]/[2^(x+3)+5]的值域 y=[(2^x)-4]/[2^(x+3)+5]=1/8*[(2^(x+3)-32)/(2^(x+3)+5)]若令 t=2^(x+3),则t&0,y=1/8*[(t-32)/(t+5)]=1/8[1-37/(t+5)]因t&0,则t+5&5,1/5&1/(t+5)&0,1-37/5&1-37/(t+5)&1所以,y的取值为大于1/8*(1-37/5)=-4/5而小于1/8.
怎么没说X的定义域
分离常数法,原是可以分离为1/8-(37/8)/[2^(x+3)+5]由此可以判断函数的单调性为单调递增的,因为2^(x+3)的值域为(0,+∝)所以可得。
顺便说一声,因为分式的定义域求法只需要分母不为0就可以了,而显然该式的分母在R上是不为0的,所以该式的定义域为R,不需要。(x-5)2+(8-x)2=15,求(x-5)(8-x)的值(两种方法)1(x-5)^2+(8-x)^2=15得到x^2-10x+25+x^2-16x+64=15得到2x^2-26x+74=0得到x^2-13x=-37(x-5)(8-x)=13x-x^2-40=37-40=-32设a=x-5,b=8-x那么a*a+b*b=15a+b=3得到(a+b)^2=9a*a+b*b+2ab=9两式相减2ab=-6ab=-3得证3((x-5)+(8-x))^2=3^2=9得到(x-5)^2+(8-x)^2+2(x-5)(8-x)=9代入得到15+2(x-5)(8-x)=9(x-5)(8-x)=(9-15)÷2=-3
你这个题有问题,你写对没?x(x-3)(2-x)(x+1)&0的解集为( ) A.(-1,1) B. C. Dx(x-3)(2-x)(x+1)&gt.hiphotos
解答此题://b.com/zhidao/pic/em/c75cfbf2df6e6092b37eca。故选B.com/zhidao/wh%3D600%2C800/sign=645c2ed18ee8acb2fd60/c75cfbf2df6e6092b37eca。 ,x(x-3)(2-x)(x+1)&gt://b.com/zhidao/wh%3D600%2C800/sign=224ba834caea15ce41bbe80f863016cb/a08b87df2678dbb91c30e924b899f336.baidu.jpg" target="_blank" title="点击查看大图" class="ikqb_img_alink"&也可利用特殊值代入检验的方法,结合下图可知
思远042zFh | 四级 采纳率72%
擅长: 暂未定制
其他类似问题
不等式(x+3)。{3x+y=0 {2(x+5)-3y=-8 {2x-3y=11 {-x+4(y-8)=7 麻烦各位大哥。3x+y=0 (1)2(x+5)-3y=-8 2x-3y=-18 (2)(1)*3+(2)得11x=-18x=-18/11代入(1)得y=54/112x-3y=11 (1)-x+4(y-8)=7 4y-x=39 (2)(1)+(2)*2得5y=89y=89/5代入 (2)得x=161/5
{2(x+5)-3y=-8{-x+4(y-8)=7,整理得{2x-3y=-18,① {x=4y-39,②把②代入①,2(4y-39)-3y=-18,5y=60,y=12,代入②,x=9.∴{x=9, {y=12.另一题。
先标号①*3+②得9x+2x=11解得x=1把x=1代入①,解得{x=1,y=-3先化简2x-3y=-18③-x+4y=39④③+2*②得5y=60解得y=12把y=12代入④。
1、y=-3x带入第二个式子2x-3*(-3x)=11x=1y=-32、{2x 10-3y=-8 {-x 4y-32=7整理{2x-3y=-18(1) {-x 4y=39(2)(2)*2 (1):-2x 8y 2x-3y=60Y=12X。
能不能难点啊:第一题:由3X+y=0 可以得出 y=-3x然后代入法代入第二个: 2x-3(-3x)=11 即11X=11 X=1 然后Y=-3第二题:由第一个弄。
X+4Y+11=0 X+Y-37=0 呼呼 我也不知道对不对啊 还是初中得时候做过 都忘差不多哒 呵呵
。解方程(1)x?32?2x+13=1;(2)4(2x+3)=8(1-x)-5(x-2);(3)1?7+3。6:1px solid black"&2。解方程组{2(x-y)-5(x+y)+29=0,x-y/7-x+y/6=-2/3 急求解答!需要。最佳答案1：亲,你好,我没有笔和纸,这里告诉你大致思路,首先把第一个方程打开化简,然后解成x=**y的形式然后把**y=x带入第二个方程,你就会明白了,纯手打,望采纳 最佳答案2：2(x-y)-5(x+y)+29=0 ---(1)x-y/7-x+y/6=-2/3 -------(2)(1)化简得3x+7y=29 -------(3)(2)化简得y=3/2代入(3)得x=37/6
我觉得你第二个式子可能写得不是很对,应该在x y和X-y加括号,才不会弄混。 先把第二个式子化简得6(x-y)-7(x y) 28=0(1),然后将第。
32+x 2 37-x 0- 5 3相关站点推荐：
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