Question

bai1:phân tích cả đa thức sau thành nhân tử

1)x^2+4y^2+4xy

2)(x-y)^2-(x-y)^2=[(x+y)-(x-y)]

3)(4x+3)^2-(2x-1)^2

4)x^3+y^3+z^3-3xy

5)x^3-2xy+y^2-z^2

Answer 1

1) \(x^2+4y^2+4xy\)

\(=\left(x+2y\right)^2\)

2 ) \(\left(x-y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\)

\(\Leftrightarrow\left(x-y-x+y\right)\left(x-y+x-y\right)=x+y-x+y\)

\(\Leftrightarrow2x-2y=2y\)

\(\Leftrightarrow2x+2y-2y=0\Leftrightarrow2\left(x+y-y\right)=0\Leftrightarrow2x=0\Leftrightarrow x=0\) :v

3 ) \(\left(4x+3\right)^2-\left(2x-1\right)^2\)

\(=\left(4x+3-2x+1\right)\left(4x+3+2x-1\right)\)

\(=\left(2x+4\right)\left(6x+2\right)\)

4 ) \(x^3+y^3+z^3-3xy\) ( thiếu đề sao á )

5 ) \(x^3-2xy+y^2-z^2\)

\(=x\left(x^2-2y\right)+\left(y-z\right)\left(y+z\right)\)

Answer 2

1) \(x^2+4y^2+4xy=x^2+\left(2y\right)^2+2.x.2y=\left(x+2y\right)^2\)

2) wtf?

3) \(\left(4x+3\right)^2-\left(2x-1\right)^2=\left(4x+3-2x+1\right)\left(4x+3+2x-1\right)\)

\(=\left(2x+4\right)\left(6x+2\right)\)

4) \(x^3+y^3+z^3-3xyz\)

\(=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

Answer 3

5) sửa đề nha bn: \(x^2-2xy+y^2-z^2\)

\(=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)