Ta có : a mũ chẵn \(\ge\)0.

=>\(2\times y-8=0\)

=> 2 x y = 8

=> y = 4

Ta có : 2x-y = 0.

=> 2x=y=8

=>x= 4

x = 2 nhé.

mk biet roi 

\(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(x.y^2\right)^6.x^3\)

\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)

\(=x^{33}.y^{31}\)

Ta có: \(x^2\cdot\left(x^4+25\right)\cdot\left(x^2-5\right)\cdot\left(x^2+5\right)\cdot\left(x-y\right)\left(x^2+xy+y^2\right)\cdot\left(x^3+y^3\right)\)

\(=x^2\cdot\left(x^4+25\right)\left(x^4-25\right)\cdot\left(x^3-y^3\right)\left(x^3+y^3\right)\)

\(=x^2\cdot\left(x^8-625\right)\cdot\left(x^6-y^6\right)\)

a, ( x - 7) ×( y+2) =0

Xét \(\left\{\begin{matrix}x-7\ge0\\y+2\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x-7=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=7\\y=-2\end{matrix}\right.\)

Vậy x= 7 và y= -2

b, ( x+2) ×(y-1)=3

Xét bảng:

Vậy cặp (x; y) thoản mãn là (-3;2); (-1;4)

c, ( 3 -x ) × ( x×y +5 ) = -1

Làm tương tự câu b

d, | x - 1| × |y+1|=2

Xét bảng:

Vậy.......

\(\dfrac{5}{6}-\dfrac{1}{2}\left(x-\dfrac{1}{3}\right)-\dfrac{2}{5}x=0\Rightarrow\dfrac{1}{2}\left(x-\dfrac{1}{3}\right)-\dfrac{2}{5}x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{6}-\dfrac{2}{5}x=\dfrac{5}{6}\Rightarrow\dfrac{1}{2}x-\dfrac{2}{5}x=\dfrac{5}{6}+\dfrac{1}{6}=1\)

\(\Rightarrow x\left(\dfrac{1}{2}-\dfrac{2}{5}\right)=1\Rightarrow\dfrac{1}{10}x=1\Rightarrow x=1:\dfrac{1}{10}=10\)

Vậy x = 10

\(\frac{x^5y}{xy^4}=\frac{x^4}{y^3}\)

\(\frac{3\times x^2\times y^5}{9\times x\times y^4}=\frac{xy}{3}\)