help me , giúp mình với 

7 nha  

HT

x + 2 x y + y = 7

x + 3 x y = 7

nếu y = 1 thì x =4

nếu y = 0 thì x = 7

nếu y = 2 thì x = 1

\(\Leftrightarrow xy=63\)

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

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: 

\(x^5=y^5\)

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

\(\Rightarrow x-y=0\)

\(\Rightarrow x=y\)

theo đầu bài ta có hệ:

\(\hept{\begin{cases}x\left(x+y+z\right)=-5\left(1\right)\\y\left(x+y+z\right)=9\left(2\right)\\z\left(x+y+z\right)=5\left(3\right)\end{cases}}\)

giải hệ ta được:\(\hept{\begin{cases}x=\frac{5}{3}\\y=-3\\z=-\frac{5}{3}\end{cases}}\)hoặc \(\hept{\begin{cases}x=-\frac{5}{3}\\y=3\\z=\frac{5}{3}\end{cases}}\)

ông Thắng giải ra luôn cho ng ta đi,cứ hệ,hệ hoài
 

Theo bài ra ta có:

\(\hept{\begin{cases}x\left(x+y+z\right)=-5\left(1\right)\\y\left(x+y+z\right)=9\left(2\right)\\z\left(x+y+z\right)=5\left(3\right)\end{cases}}\)

Cộng từng vế các đẳng thức

\(x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=-5+9+5=9\)

\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=9\Rightarrow\left(x+y+z\right)^2=9\)

\(\Rightarrow\hept{\begin{cases}x+y+z=3\\x+y+z=-3\end{cases}}\)

TH1:x+y+z=3

Từ x(x+y+z)=-5

=>\(x=x\left(x+y+z\right):\left(x+y+z\right)=-5:3\Rightarrow x=-\frac{5}{3}\)

Từ y(x+y+z)=9

=>\(y=y\left(x+y+z\right):\left(x+y+z\right)=9:3=3\)

Từ z(x+y+z)=5

=>\(z=z\left(x+y+z\right):\left(x+y+z\right)=5:3=\frac{5}{3}\)

TH2:giải tương tự

Đ/S: \(\left(x;y;z\right)\in\left\{\left(-\frac{5}{3};3;\frac{5}{5}\right);\left(\frac{5}{3};-3;-\frac{5}{3}\right)\right\}\)

\(\Rightarrow x\left(2y+1\right)-3\left(2y+1\right)=7\)

\(\Leftrightarrow\left(x-3\right)\left(2y+1\right)=7=1.7=7.1=-1.-7=-7.-1\)

vậy....

cảm ơn bạn Nguyễn Xuân Anh rất nhiều

12345(6787000)=-506

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

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