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a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
a, x(a - b) + (a - b)
= (x + 1)(a - b)
b, x(a + b) - a - b
= x(a + b) - (a + b)
= (x - 1)(a + b)
c, 10ax - 5ay - 2x + y
= 5a(2x - y) - (2x - y)
= (5a - 1)(2x - y)
d, 2a^2x - 5by - 5a^2y + 2bx
= 2x(a^2 + b) - 5y(b + a^2)
= (2a - 5y)(a^2 + b)
làm tiếp:
2ax2 - bx2 - 2ax +bx +4a-2b
= x2(2a-b) - x(2a-b) +2(2a-b)
=(2a-b)(x2-x+2)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(2x^2y+4xy^2+2y^3-8\)
\(=2y\left(x^2+2xy+y^2\right)-8\)
\(=2y\left(x+y\right)^2-8\)
\(=2\left[y\left(x+y\right)^2-4\right]\)
\(2x^2+2y^2-4xy-xz+yz=\left(2x^2+2y^2-4xy\right)-\left(xz-yz\right)=2\left(x^2-2xy+y^2\right)-z\left(x-y\right)=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\)
\(=2\left(x^2-2xy+y^2\right)-\left(xz-yz\right)\)
\(=\left(x-y\right)\left(2x-2y-z\right)\)