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Ta có: x2y+5xy-14y
= y(x2+5x-14)
= y(x+7)(x-2)
x3-5x2-14x
= x(x2-5x-14)
= x(x-7)(x+2)
\(x^2y+5xy-14y\)
\(=x^2y-2xy+7xy-14y\)
\(=xy\left(x-2\right)+7y\left(x-2\right)\)
\(=\left(x-2\right)\left(xy+7y\right)\)
\(x^3-5x^2-14x\)
\(=x^3-7x^2+2x^2-14x\)
\(=x^2\left(x-7\right)+2x\left(x-7\right)\)
\(=\left(x^2+2x\right)\left(x-7\right)\)
\(=x\left(x+2\right)\left(x-7\right)\)
\(x^2-y^2-5x+5y=\left(x^2-y^2\right)-\left(5x-5y\right)\)
\(=\left(x+y\right)\left(x-y\right)-5\cdot\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left[\left(x+y\right)-5\right]=\left(x-y\right)\cdot\left(x+y-5\right)\)
ko ghi lại đề nha:
=(5x2-10x)+5y2-5z
=5(x-2)+5y2-5z
mk chỉ làm đc đến đây thôi
x2 - 5x + 5y - y2 =
= -5 (x-y) + (x2 - y2 )
= -5 (x-y) + (x-y)(x+y)
= (x-y) (-5 +x+y)
x2-5x+5y-y2
=x2-y2-5x+5y
=(x2-y2)-5(x+y)
=(x2-y2)(-5+x+y)
\(5x^2+3\left(x+y\right)^2-5y^2\)
\(=5x^2-5y^2+3\left(x+y\right)^2\)
\(=5\left(x^2-y^2\right)+3\left(x+y\right)^2\)
\(=5\left(x+y\right)\left(x-y\right)+3\left(x+y\right)^2\)
\(=\left(x+y\right)\left[5\left(x-y\right)+3\left(x+y\right)\right]\)
\(=\left(x+y\right)\left[5x-5y+3x+3y\right]\)
\(=\left(x+y\right)\left(8x-2y\right)\)
\(=2\left(x+y\right)\left(4x-y\right)\)
\(5x^3-5xy^2+10xy-5x\)
\(=5x\left(x^2-y^2+2y-1\right)\)
\(=5x\left[x^2-y^2+y+y-1\right]\)
\(=5x\left[x^2-y\left(y-1\right)+\left(y-1\right)\right]\)
\(=5x\left[x^2-\left(y+1\right)\left(y-1\right)\right]\)
\(=5x\left[x^2-\left(y^2-1^2\right)\right]\)
\(=5x\left[\left(x-y\right)\left(x+y\right)+1\right]\)
Lâu k làm, sai thông cảm
a/ 5x2-x+5xy-y
= 5x(x+y) - (x+y)
= (5x-1) (x+y)
b) 2xz + 3z + 6y + xz
= ko bt làm
a; x(5x-1)+y(5x-1)
=>( x+y).(5x-1)
b; có ghi sai đề ko vậy ; toàn x vs z thế
5x^3y + 5x^2y - 5xy - 5y
=5y(x^3+x^2-x-1)
=5y(x^2(x+1)-(x+1))
=5y(x^2-1)(x+1)
=5y(x-1)(x+1)^2
=5y(x-1)(x^2+2x+1)
5x3y+5x2y-5xy-5y=5y(x3+x2-x-1)
=5y[x2(x+1)-(x+1)]=5y[(x2-1)(x+1)]=5y(x-1)(x+1)2