Phân tích đa thức thành nhân tử
x3+7x2y+7xy2+y3
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\(=x^3-x+7x+7=x\left(x-1\right)\left(x+1\right)+7\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x+7\right)\)
Sửa đề: x^3+6x^2+11x+6
=x^3+x^2+5x^2+5x+6x+6
=(x+1)(x^2+5x+6)
=(x+1)(x+2)(x+3)
a) \(=2y^3\left(x^2-16\right)=2y^3\left(x-4\right)\left(x+4\right)\)
b) \(=7y\left(x^2-2x+1\right)=7y\left(x-1\right)^2\)
c) \(=2x^2\left(x+5y\right)-y\left(x+5y\right)=\left(x+5y\right)\left(2x^2-y\right)\)
a: \(2x^2y^3-32y^3=2y^3\left(x-4\right)\left(x+4\right)\)
b: \(7x^2y-14xy+7y=7y\left(x^2-2x+1\right)=7y\left(x-1\right)^2\)
Câu 1:
$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$
$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$
Câu 2:
$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$
$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$
Câu 1:
\(x^2+4y^2+4xy-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
Câu 2:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)
\(x^3-y^3+2x^2+2xy\)
\(=x\left(x^2-y^2+2x+2y\right)\)
\(=\)\(x\left[\left(x+y\right)\left(x-y\right)+2\left(x+y\right)\right]\)
\(=x\left(x+y\right)\left(x-y+2\right)\)
`(x+y)^3-x^3-y^3`
`=(x+y)^3-(x^3+y^3)`
`=(x+y)^3-(x+y)(x^2-xy+y^2)`
`=(x+y)[(x+y)^2-x^2+xy-y^2]`
`=(x+y)(x^2+2xy+y^2-x^2+xy-y^2)`
`=(x+y).3xy`
a) Ta có: \(\left(x+y\right)^3-x^3-y^3\)
\(=x^3-x^3+y^3-y^3+3x^2y+3xy^2\)
\(=3xy\left(x+y\right)\)
\(x^3+y^3+3y^2+3y+1\\ =x^3+\left(y+1\right)^3\\ =\left(x+y+1\right)\left[x^2-x\left(y+1\right)+\left(y+1\right)^2\right]\\ =\left(x+y+1\right)\left(x^2-xy-x+y^2+2y+1\right)\\ =\left(x+y+1\right)\left(x^2+y^2+2y+1-xy-x\right)\)
\(x^3+7x^2y+7xy^2+y^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3+4x^2y+4xy^2\)
\(\Leftrightarrow\left(x+y\right)^3+4xy\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2+4xy\right]\)
\(\Leftrightarrow\left(x+y\right)\left(x^2+6xy+y^2\right)\)
P/s tham khảo nha
\(x^3+7x^2y+7xy^2+y^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3+4x^2y+4xy^2\)
\(\Leftrightarrow\left(x+y\right)^3+4xy\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2+4xy\right]\)
\(\Leftrightarrow\left(x+y\right)\left(x^2+6xy+y^2\right)\)
P/s tham khảo nha