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AH
Akai Haruma
Giáo viên
2 tháng 3 2020
Lời giải:
ĐK: $x\neq 5;x\neq 0; y\neq 2; y\neq -1$
\(M=\frac{x^2-25}{x^3-10x^2+25x}:\frac{y-2}{(y-2)(y+1)}=\frac{(x-5)(x+5)}{x(x^2-10x+25)}:\frac{1}{y+1}\)
\(=\frac{(x-5)(x+5)}{x(x-5)^2}:\frac{1}{y+1}=\frac{x+5}{x(x-5)}.(y+1)=\frac{(x+5)(y+1)}{x(x-5)}\)
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$x^2+9y^2-4xy=2xy-|x-3|$
$\Leftrightarrow x^2+9y^2-6xy=-|x-3|$
$\Leftrightarrow (x-3y)^2+|x-3|=0$
Dễ thấy $(x-3y)^2\geq 0; |x-3|\geq 0$ với mọi $x,y\in $ĐKXĐ nên để tổng của chúng bằng $0$ thì:
$x-3y=x-3=0\Rightarrow x=3; y=1$
Khi đó: $M=\frac{(3+5)(1+1)}{3(3-5)}=\frac{-8}{3}$
LT
14 tháng 10 2017
a. 6x3y2 ( 2-x) + 9x2y2 (x-2)
= -6x3y2 (x-2) + 9x2y2 ( x-2)
= (x-2) 3x2y2 ( -2x + 3)
b. x2 - 4x + 4y - y2
= x2 - y2 - (4x - 4y )
= (x-y)(x+y) - 4( x-y)
= (x-y)(x+y-4)
c. 81x2 + 6yz -9y2-z2
= 81x2 - (9y2 - 6yz + z2 )
= (9x)2 - ( 3y - z )2
= (9x + 3y -z)(9x - 3y + z )
14 tháng 10 2017
\(a,=6x^3y^2\left(2-x\right)-9x^2y^2\left(2-x\right)\)
\(=\left(2-x\right)\left(6x^3y^2-9x^2y^2\right)=\left(2-x\right)3x^2y^2\left(2x-3\right)\)
\(f,=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5\left(2x-1\right)\)
\(g,=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x-3+2\right)\)
\(=\left(x+3\right)\left(x-1\right)\)
28 tháng 1 2020
c) \(x^2+x-ax-a\)
\(=x\left(x+1\right)-a\left(x+1\right)\)
\(=\left(x+1\right)\left(x-a\right)\)
d) \(2xy-ax+x^2-2ay\)
\(=2y\left(x-a\right)+x\left(x-a\right)\)
\(=\left(x-a\right)\left(2y+x\right)\)
e) \(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
f) \(25-10x-4y^2+x^2\)
\(=\left(x^2-10x+25\right)-\left(2y\right)^2\)
\(=\left(x-5\right)^2-\left(2y\right)^2\)
\(=\left(x-5-2y\right)\left(x-5+2y\right)\)
g) \(x^3-6xy+9y^2-36\)
h) \(4x^2-9y^2+4x-6y\)
\(=\left(2x\right)^2-\left(3y\right)^2+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
k) \(-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(-x+y+5\right)\)
i) \(4x^2-25y^2-6x+15y\)
\(=\left(2x\right)^2-\left(5y\right)^2-3\left(2x-5y\right)\)
\(=\left(2x-5y\right)\left(2x+5y\right)-3\left(2x-5y\right)\)
\(=\left(2x-5y\right)\left(2x+5y-3\right)\)
KD
28 tháng 1 2020
a, \(x\left(y+z\right)^2+y\left(x+z\right)^2+z\left(x+y\right)^2+4xyz\)
\(=x\left(y+z\right)^2+x^2\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(xy+xz+z^2+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b, \(yz\left(y+z\right)+xz\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+xz^2-x^2z-x^2y-xy^2\)
\(=yz\left(y+z\right)-x\left(y+z\right)\left(y-z\right)-x^2\left(y+z\right)\)
\(=\left(y+z\right)\left(yz-xy+xz-x^2\right)\)
\(=\left(y+z\right)\left[y\left(z-x\right)+x\left(z-x\right)\right]\)
\(=\left(y+z\right)\left(y+x\right)\left(z-x\right)\)
LT
1 tháng 8 2020
a) \(2x^2y^2-\frac{4}{3}x^2y+2xy\)
\(=xy\left(2xy-\frac{4}{3}x+2\right)\)
b) 2xy2.(x + 5y) - 4xy(5y + x)
= (5y + x)(2xy2 - 4xy)
= 2xy(5y + x)(y - 2)
c) 25 - 4x2 - y2 + 4xy
= 25 - (4x2 - 4xy + y2)
= 52 - (2x + y)2
= (5 - 2x - y)(5 + 2x + y)
d) x2 + 4x - 2xy - 4y +y2
= (x2 - 2xy + y2) + (4x - 4y)
= (x - y)2 + 4(x - y)
= (x - y)(x - y + 4)
e) 12y3 - 3x2y + 12xy - 12y
= 3y(4y2 - x2 + 4x - 4)
= 3y[4y2 - (x - 2)2]
= 3y(2y - x + 2)(2y + x - 2)
f) 64x4 + y4
= (8x2)2 + 16x2y2 + y4 - 16x2y2
= (8x2 + y2)2 - (4xy)2
= (8x2 + y2 - 4xy)(8x2 + y2 + 4xy)
TG
1 tháng 8 2020
a) \(2x^2y^2-\frac{4}{3}x^2y+2xy\)
b) \(2xy^2\left(x+5y\right)-4xy\left(5y+x\right)\)
\(=\left(x+5y\right)\left(2xy^2-4xy\right)\)
\(=2\left(x+5y\right)\left(xy^2-2xy\right)\)
c) \(25-4x^2-y^2+4xy\)
\(=25-\left(4x^2+y^2-4xy\right)\)
\(=5^2-\left[\left(2x\right)^2-2.2x.y+y^2\right]\)
\(=5^2-\left(2x-y\right)^2\)
\(=\left(5-2x+y\right)\left(5+2x-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y\right)+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(12y^3-3x^2y+12xy-12y\)
f) \(64x^4+y^4\)
\(=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)
23 tháng 5 2022
1: \(=x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3+y\right)\left(x+3-y\right)\)
2: \(x^2-2xy+y^2-25\)
\(=\left(x-y\right)^2-25\)
\(=\left(x-5-y\right)\left(x+5-y\right)\)
4: \(=y\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(y-5\right)\)
5: \(=x^3\left(x+3\right)-9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^3-9\right)\)
4 tháng 8 2019
a) 10x(x - y)2 - 5(x - y)3 = [10x - 5(x - y)](x - y)2 = (10x - 5x + y)(x - y)2 = (5x + y)(x - y)2
b) -x2 - 10x - 25 = -(x2 + 10x + 52) = -(x + 5)2
c) 64x6y4 - 81x2y2 = (8x3y2)2 - (9xy)2 = (8x3y2 + 9xy)(8x3y2 - 9xy)
d) x6 - y6 = (x3)2 - (y3)2 = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)
e)1/8x3 - 3/4x2y + 3/2xy2 - y3 = (1/2x)3 - 3.(1/2x)2y + 3.1/2xy2 - y3 = (1/2x - y)3
f) (3x + 1)2 - (x - 1)2 = (3x + 1 + x - 1)(3x + 1 - x + 1) = 4x(2x + 2) = 8x(x + 1)
28 tháng 1 2020
Bài 1:
a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
c)Đề sai hoàn toàn
d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)
f)\(\left(x^2+y^2\right)^2-4x^2y^2=\left(x-y\right)^2\left(x+y\right)^2\)
g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)
i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)