Phan tich thanh nhan tu
x^2 - y - y^2 - x
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\(=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
\(x^2+y^2-x^2y^2+xy-x-y\)\(=\left(xy-x\right)+\left(y^2-y\right)-\left(x^2y^2-x^2\right)\)
\(=x\left(y-1\right)+y\left(y-1\right)-x^2\left(y^2-1\right)\)\(=\left(y-1\right)\left[x+y-x^2\left(y+1\right)\right]\)
\(=\left(y-1\right)\left(x+y-x^2y-x^2\right)\)\(=\left(y-1\right)\left[x\left(1-x\right)+y\left(1-x^2\right)\right]\)
\(=\left(y-1\right)\left(1-x\right)\left[x+y\left(1+x\right)\right]=\left(y-1\right)\left(1-x\right)\left(xy+x+y\right)\)
Ta có:
\(5x^2+3\left(x+y\right)^2-5y^2\)
\(\Rightarrow5x^2+3.\left(x^2+2xy+y^2\right)-5y^2\)
\(\Rightarrow5x^2+3x^2+6xy+3y^2-5y^2\)
\(\Rightarrow8x^2+6xy-2y^2\)
x2 - 3x + 3y - y2
= (x2 - y2) - (3x - 3y)
= (x - y)(x + y) - 3(x - y)
= (x - y)(x + y - 3)
= x2 - y2 - 3x+3y = (x-y)(x+y) -3(x-y)
= (x+y+3)(x-y)
nhớ chọn cho mk nha!!!!!!
\(x^2-y-y^2-x=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
x2 - y - y2 - x
= (x2 - y2) - x - y
= (x - y)(x + y) - (x + y)
= (x - y - 1)(x + y)