phân tích đa thức thành nhân tử
x^2-5x+2xy-10y
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=(x^5-5x^2)-(2xy+10y)
=(x^5-5x^2)-(2xy-10y)
=x^2.(x-5)-2y.(x-5)
=(x^2-2y).(x-5)
k nha
\(x^3-5x^2-2xy+10y=x^2\left(x-5\right)-2y\left(x-5\right)=\left(x-5\right)\left(x^2-2y\right)=\left(x-5\right)\left(x-2y\right)\left(x+2y\right)\)
\(=2xy+5x+4y^2+10y\)
\(=x\left(2y+5\right)+2y\left(2y+5\right)\)
\(=\left(x+2y\right)\left(2y+5\right)\)
\(=2x\left(x+2y\right)+5\left(x+2y\right)=\left(x+2y\right)\left(2x+5\right)\)
\(=x^2y\left(x-5\right)-2y\left(x-5\right)+0\)
\(=\left(x-5\right)\left(x^2y-2y\right)\)
xong phân tích nốt cái bậc 2 kia để được max điểm :)))
a: x^2+4xy-21y^2
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
=5x(x+y)+y(x+y)
=(x+y)(5x+y)
c: \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
=x(x+5y)-3y(x+5y)
=(x+5y)(x-3y)
d: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
=x(x-2y)-5y(x-2y)
=(x-2y)(x-5y)
a) \(x^2+4xy-21y^2\)
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b) \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(5x+y\right)\left(x+y\right)\)
c) \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
\(=x\left(x+5y\right)-3y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
d) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)
\(x^2-2xy+5x-10y\)
\(=x\left(x-2y\right)+5\left(x-2y\right)\)
\(=\left(x+5\right)\left(x-2y\right)\)
\(x^2-2xy+5x-10y\)
\(=\left(x^2-2xy\right)+\left(5x-10y\right)\)
\(=x\left(x-2y\right)+5\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+5\right)\)
\(x-3\sqrt{x}+\sqrt{xy}-3y\)
\(=\left(x-3\sqrt{x}\right)+\left(\sqrt{xy}-3y\right)\)
\(=\sqrt{x}\left(\sqrt{x}-3\right)+y\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right)\left(\sqrt{x}+y\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(x^2-5x+2xy-10y\)
\(=\left(x^2-5x\right)+\left(2xy-10y\right)\)
\(=x\left(x-5\right)+2y\left(x-5\right)\)
\(=\left(x+2y\right)\left(x-5\right)\)