\(\begin{array}{l}\left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right) + \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\ = {x^3} - {\left( {2y} \right)^3} + {x^3} + {\left( {2y} \right)^3}\\ = {x^3} - 8{y^3} + {x^3} + 8{y^3}\\ = 2{x^3}\end{array}\)

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

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

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

$\Leftrightarrow x^2+2=y$

https://olm.vn/hoi-dap/detail/108858274535.html

Bài tương tự gưi link ib

\(\hept{\begin{cases}\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\\\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\end{cases}}\)

<=> \(\hept{\begin{cases}x^3+8y^3=0\left(1\right)\\x^3-8y^3=16\left(2\right)\end{cases}}\)

Lấy (1) + (2) theo vế

=> 2x³ = 16

=> x³ = 8 = 2³

=> x = 2

Thế x = 2 vào (1)

=> 2³ + 8y³ = 0

=> 8 + 8y³ = 0

=> 8y³ = -8

=> y³ = -1 = (-1)³

=> y = -1

Vậy \(\hept{\begin{cases}x=2\\y=-1\end{cases}}\)

Từ \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\)

\(\Leftrightarrow x^3+\left(2y\right)^3=0\)\(\Leftrightarrow x^3+8y^3=0\)(1)

Từ \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\)

\(\Leftrightarrow x^3-\left(2y\right)^3=16\)\(\Leftrightarrow x^3-8y^3=16\)(2)

Cộng (1) với (2) ta được: \(\left(x^3+8y^3\right)+\left(x^3-8y^3\right)=16\)

\(\Leftrightarrow2x^3=16\)\(\Leftrightarrow x^3=8\)\(\Leftrightarrow x=2\)

Thay \(x=2\)vào (1) ta được: 

\(2^3+8y^3=0\)\(\Leftrightarrow8y^3+8=0\)

\(\Leftrightarrow8y^3=-8\)\(\Leftrightarrow y^3=-1\)\(\Leftrightarrow y=-1\)

Vậy \(x=2\); \(y=-1\)

a.

Với \(y=0\) không phải nghiệm

Với \(y\ne0\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5}{y}\\2x\left(x+y\right)+y=\dfrac{5}{y}\end{matrix}\right.\)

\(\Rightarrow3x+2=2x\left(x+y\right)+y\)

\(\Leftrightarrow2x^2+\left(2y-3\right)x+y-2=0\)

\(\Delta=\left(2y-3\right)^2-8\left(y-2\right)=\left(2y-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2y+3+2y-5}{4}=-\dfrac{1}{2}\\x=\dfrac{-2y+3-2y+5}{4}=-y+2\end{matrix}\right.\)

Thế vào pt đầu ...

Câu b chắc chắn đề sai

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+8y^3=0\\x^3-8y^3=16\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^3=8\\y^3=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)