ta có : `x/2 = y/3 = z/4=> (2x)/4 =(3y)/9 = z/4`

`=> (2x)/4 =(3y)/9 = z/4` và `2x + 3y - z = 27`

Áp dụng t/c dãy tỉ số bằng nhau ta có:

`(2x)/4 =(3y)/9 = z/4 =(2x + 3y - z)/(4+9-4)=27/9=3`

`=>x/2=3=>x=3.2=6`

`=>y/3=3=>x=3.3=9`

`=>z/4=3=>z=3.4=12`

\(\Leftrightarrow xy=63\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)

Ta có: 

\(x^4=y^4\)

\(\Rightarrow x^4-y^4=0\)

\(\Rightarrow\left(x^2\right)^2-\left(y^2\right)^2=0\)

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

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

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

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

_______________

Ta có: 

\(x^5=y^5\)

\(\Rightarrow x^5-y^5=0\)

\(\Rightarrow x-y=0\)

\(\Rightarrow x=y\)

a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)

Thay x-y=7 vào biểu thức (1), ta được:

\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)

Vậy: Khi x-y=7 thì A=100

b) Ta có: \(x+y=2\)

\(\Leftrightarrow\left(x+y\right)^2=4\)

\(\Leftrightarrow x^2+y^2+2xy=4\)

\(\Leftrightarrow2xy+10=4\)

\(\Leftrightarrow2xy=-6\)

\(\Leftrightarrow xy=-3\)

Ta có: \(A=x^3+y^3\)

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

Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:

\(A=2\cdot\left(10+3\right)=2\cdot13=26\)

Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26

\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)

\(\dfrac{x}{3}=x+y=20\Rightarrow x=60\Rightarrow60+y=20\Rightarrow y=-40\)

Ta có:

\(\dfrac{x}{3}=20\)

\(\Rightarrow\)\(x=60\)

Lại có:

\(x+y=20\)

\(\Rightarrow\)\(y=20-60\)

\(\Rightarrow\)\(y=-40\)

Vây x = 60 và y = - 40

\(\dfrac{x}{3}=x+y\)

\(\Leftrightarrow x-\dfrac{1}{3}x=-y\)

\(\Leftrightarrow y=-\dfrac{2}{3}x\)

Ta có: x+y=20

\(\Leftrightarrow x\cdot\dfrac{1}{3}=20\)

hay x=60

=> y=40

Ta có: \(x+y+z=0\)

nên \(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)

Ta có: \(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

\(=\dfrac{x+y}{y}\cdot\dfrac{y+z}{z}\cdot\dfrac{x+z}{x}\)

\(=\dfrac{-z}{y}\cdot\dfrac{-x}{z}\cdot\dfrac{-y}{x}\)

\(=\dfrac{-\left(x\cdot y\cdot z\right)}{x\cdot y\cdot z}=-1\)