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a) Ta có: \(x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
b) Ta có: \(2x+2y-x^2-xy\)
\(=2\left(x+y\right)-x\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x\right)\)
c) Ta có: \(x^2-25+y^2+2xy\)
\(=\left(x+y\right)^2-25\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
d) Ta có: \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
e) Ta có: \(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
f) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\(A=4x^2+12xy+9y^2\)
\(B=25x^2-10xy+y^2\)
\(C=8x^3+12x^2y^2+6xy^4+y^6\)
\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)
\(E=x^3-27y^3\)
\(F=x^6-27\)
, \(B=\frac{2x^2+4xy}{y^2+z^2}=\frac{2x\left(x+2y\right)}{y^2+z^2}\)
\(\hept{\begin{cases}x-y-z=0\\x+2y-10z=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-y=z\\x+2y=10z\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4z\\y=3z\end{cases}}\)
Thay vào B, ta được: \(B=\frac{2.\left(4z\right)^2+4.4z.3z}{\left(3z\right)^2+z^2}=\frac{2.4^2+3.4^2}{3^2+1}=8\)
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