`a, (x-y)(x-5y)`

`= x^2 - xy - 5xy + 5y^2`

`= x^2 - 6xy + 5y^2`

`b, (2x+y)(4x^2 -2xy + y^2)`

`= (2x)^3 + y^3`

`= 8x^3 + y^3`

a) \(\left(x-y\right)\left(x-5y\right)\)

\(=x^2-5xy-xy+5y^2\)

\(=x^2-6xy+5y^2\)

b) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

\(=8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3\)

\(=8x^3+y^3\)

a) \(\left(-5a^4\right)\cdot\left(a^2b-ab^2\right)\)

\(=\left(-5a^4\cdot a^2b\right)-\left(-5a^4\cdot ab^2\right)\)

\(=-5a^6b+5a^5b^2\)

b) \(\left(x+2y\right)\left(xy^2-2y^3\right)\)

\(=x^2y^2-2xy^3+2xy^3-4y^4\)

\(=x^2y^2-4y^4\)

`a, (-5a^4)(a^2b - ab^2)`

`= -5(a^(4+2) . b) + 5a^(4+1) . b^2`

`= -5a^6b + 5a^5b^2`

`b, (x+2y)(xy^2-2y^3)`

`= x^2y^2 + 2xy^3 - 2xy^3 - 4y^4`

b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu

c)

ĐK $y \geqslant 0$

Hệ đã cho tương đương với

$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$

Trừ từng vế $2$ phương trình ta được

$x^2+2+2\sqrt{y(x^2+2)}-3y=0$

$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$

$\Leftrightarrow x^2+2=y$

`a, (4-x)(4+x) = 16 - x^2`

`b, (2y+7z)(2y-7z) = 4y^2 - 49z^2`

`c, (x+2y^2)(x-2y^2)`

`= x^2 - 4y^4`

a) ĐKXĐ: \(x;y\ne0,x\ne\frac{y}{2},y\ne\frac{x}{2}\)
\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}\)\(=\frac{y^2-4x^2}{xy\left(2x-y\right)}=\frac{\left(y-2x\right)\left(y+2x\right)}{xy\left(2x-y\right)}\)
\(=\frac{-\left(y+2x\right)}{xy}\)

b) ĐKXĐ: \(x\ne2;x\ne-2\)
\(\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)\(=\frac{1}{x+2}+\frac{3}{\left(x-2\right)\left(x+2\right)}+\frac{x-14}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x^2+4x+4\right)-16}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x+2\right)^2-16}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x+2-4\right)\left(x+2+4\right)}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\frac{x+6}{\left(x+2\right)^2}\)