\(A=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\left(x+y\right)\)

\(=1-3xy+3xy\left[1-2xy\right]+6x^2y^2\)

=1

a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

=>x+1=1 và y-2=1/2

=>x=0 và y=5/2

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)

=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6

=>x-2y=9 và 2x-y=12

=>x=5; y=-2

c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)

=>|x-6|=1 và |y+1|=1

=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)

làm họ mình đi

huhu co to chua day

giải dc ùi

2/

2(x⁶+y⁶)-3(x⁴+y⁴)

=2[(x²)³+(y²)³ ] - 3x⁴-3y⁴

=2(x²+y²)(x⁴-x²y²+y⁴)-3x⁴-3y⁴

=2.1(x⁴-x²y²+y⁴)-3x⁴-3y⁴

=2x⁴-2x²y²+2y⁴-3x⁴-3y⁴

=-x⁴-2x²y²-y⁴

=-(x⁴+2x²y²+y⁴)

=-(x²+y²)

=-1