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5x^2-5xy-10x+10y
(5x^2-5xy )-(10x-10y)
5x(x-y)-10(x-y)
(x-y)(5x-10)
3(x2-2xy+y2):10(x-y)
3(x-y)2:10(x-y)
3(x-y):10
5(x2-xy+y2):(x+y)(x2-xy+y2)
5:x+y
ở nơi 5y thiếu ^2 nhé
Bài 1:
a) x( x - y) + x - y = (x - y)(x + 1)
b) 2x + 2y - x( x + y) = ( 2x + 2y) - x( x + y)
= 2( x + y ) - x( x + y ) = ( x + y )(2 - x )
c) 5x2 - 5xy - 10x + 10y = ( 5x2 - 5xy ) - ( 10x - 10y)
= 5x( x - y ) - 10( x - y ) = ( x - y )(5x - 10 )
= 5( x - y )( x - 2 )
d) 4x2 + 6xy - 3x - 6y = Mình ko làm được!!! bạn chép có sai đề không
Bài 2:
x ( 2x - 7) - 4x + 14 = 0
⇒ 2x2 - 7x - 4x + 14 = 0 ⇒ ( 2x2 - 4x ) - ( 7x - 14 ) = 0
⇒ 2x( x - 2 ) - 7(x - 2) = 0
⇒ (x - 2)(2x - 7) = 0
⇒ \(\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x = 2; x = \(\dfrac{7}{2}\)
1) 2x + 2y - x(x+y)
= 2(x + y) - x(x + y)
= (2 - x)(x + y)
2/ 5x2 - 5xy -10x + 10y
= 5x(x - y) - 10(x - y)
= (5x - 10(x - y)
3/ 4x2 + 8xy - 3x - 6y
= 4x(x + 2y) - 3(x + 2y)
= (4x - 3)(x + 2y)
1) 2x + 2y - x(x + y)
= 2(x + y) - x(x + y)
= (2 - x)(x + y)
2) 5x2 - 5xy - 10x + 10y
= 5x(x - y) - 10(x - y)
= (5x - 10)(x - y)
= 5(x - 2)(x - y)
3) 4x2 + 8xy - 3x - 6y
= 4x(x + 2y) - 3(x + 2y)
= (4x - 3)(x + 2y)
4) 2x2 + 2y2 - x2z + z - y2z - 2
= 2(x2 + y2 - z(x2 + y2) - (2 - z)
= (2 - z)(x2 + y2) - (2 - z)
= (2 - z)(x2 + y2)
5) x2 + xy - 5x - 5y
= x(x + y) - 5(x + y)
= (x - 5)(x + y)
6) x(2x - 7) - 4x + 14
= x(2x - 7) - 2(2x - 7)
= (x - 2)(2x - 7)
7)x2 - 3x + xy - 3y
= x(x + y) - 3(x + y)
= (x - 3)(x + y)
a) \(x^2+2x+1-y^2 = (x^2 +2x+1) – y^2\)
\(= (x+1)^2 - y^2\)
\(= (x+1+ y) (x+1- y)\)
b) \(2xy+z+2x+yz\)
\(=\left(2xy+2x\right)+\left(z+yz\right)\)
\(=2x\left(y+1\right)+z\left(y+1\right)\)
\(=\left(2x+z\right)\left(y+1\right)\)
c) \(x^2+x-xy-y\)
\(=\left(x^2+x\right)-\left(xy+y\right)\)
\(=x\left(x+1\right)-y\left(x+1\right)\)
\(=\left(x-y\right)\left(x+1\right)\)
d) \(5x^2-5xy+10x+10y\)
\(=\left(5x^2-5xy\right)+\left(10x+10y\right)\)
\(=5x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-10\right)\)
a, x2+2x+1-y2= (x+1)2-y2= (x+1-y)(x+1+y)
b, 2xy+z+2x+yz= (2xy+2x)+(z+yz)
= 2x(y+1)+z(y+1)
= (2x+z)(y+1)
c, x2 +x - xy - y = x(x+1) - y(x+1)
= (x - y)(x+1)
d, 5x2 - 5xy - 10x + 10y=5x(x-y) - 10(x-y)
= (5x-10)(x-y).(Câu này có thể là sai đề nên mik đã sửa lại đề cho dễ làm nha). Chúc bạn học tốt!!!
Câu 1
2/5.xy(x2y-5x+10y)
=2/5.xy.x2y-2/5.xy.5x+2/5.xy.10y
=2/5.x3y2-2x2y+4xy2
Câu 2
3x(x2+x-1)
=3x.x2+3x.x-3x.1
=3x3+3x2-3x
a) \(x^2-5xy+2x-10y\)
\(=\left(x^2-5xy\right)+\left(2x-10y\right)\)
\(=x.\left(x-5y\right)+2.\left(x-5y\right)\)
\(=\left(x-5y\right).\left(x+2\right)\)
b) \(x^2-5x+4\)
\(=x^2-4x-x+4\)
\(=\left(x^2-4x\right)-\left(x-4\right)\)
\(=x.\left(x-4\right)-\left(x-4\right)\)
\(=\left(x-4\right).\left(x-1\right)\)
Chúc bạn học tốt!
\(5x^2-5xy-10x+10y\)
`=5x(x-y)-10(x-y)`
`=5(x-y)(x-2)`
\(5x^2-5xy-10x+10y\)
\(=5\left(x^2-xy-2x+2y\right)\)
\(=5\left[x\left(x-y\right)-2\left(x-y\right)\right]\)
\(=5\left(x-y\right)\left(x-2\right)\)