x² - 4y² + x + 2y

= ( x² - 4y² ) + ( x + 2y )

= ( x - 2y ) ( x + 2y ) + ( x + 2y )

= ( x + 2y ) ( x - 2y + 1 )

Bài 2: 

Sửa đề:  \(x^3-3x^2-10x=0\)

\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)

\(=\dfrac{2\left(x-2y+z\right)^3+4\left(x-2y+z\right)^2}{2\left(x-2y+z\right)}=\left(x-2y+z\right)^2+2\left(x-2y+z\right)\)

\(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

\(=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

\(x^2-9x-y^2-9y\)

\(=\left(x^2-y^2\right)-\left(9x+9y\right)\)

\(=\left(x-y\right)\left(x+y\right)-9\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-9\right)\)

b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)

Câu hỏi của Lee Min Ho - Toán lớp 8 - Học toán với OnlineMath

Câu hỏi của Lee Min Ho - Toán lớp 8 - Học toán với OnlineMath

x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz

=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz

=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz

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

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

=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)

=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]

=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]

=(x+y+z)(x-y-z)(z-x-y)