Check lại :v

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

1.\(=x^3+8y^3-x^3+8y^3+2y^3=18y^3\)

2. \(=x^3-3x^2+3x-1+1-x^3+3\left(9-x^2\right)\)

\(=-3x^2+3x+27-3x^2=3\left(x+9\right)\)

Ko chắc lém :))))

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

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

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

\(A=B.C\) đặt \(\left\{{}\begin{matrix}a=\sqrt{x}\\b=\sqrt{2y}\end{matrix}\right.\)

\(B=\dfrac{2a^2+b^2}{\left(a-b\right)\left(a^2+b^2+ab\right)}-\dfrac{a}{a^2+ab+b^2}\)

\(B=\dfrac{2a^2+b^2-a\left(a-b\right)}{\left(a-b\right)\left(a^2+b^2+ab\right)}=\dfrac{a^2+b^2+ab}{\left(a-b\right)\left(a^2+b^2+ab\right)}\)

\(B=\dfrac{1}{a-b}\)

\(C=\dfrac{a^3+b^3}{b^2+ab}-a=\dfrac{\left(a+b\right)\left(a^2+b^2-ab\right)}{b\left(a+b\right)}-a=\dfrac{a^2+b^2-ab-ab}{b}\)

\(C=\dfrac{\left(a-b\right)^2}{b}\)

\(A=\dfrac{1}{a-b}.\dfrac{\left(a-b\right)^2}{b}=\dfrac{a-b}{b}=\dfrac{a}{b}-1\)

\(A=\sqrt{\dfrac{x}{2y}}-1\)

A=\(\sqrt{\dfrac{x}{y2}}-1\)

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

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

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

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