`x^3+1=2y,y^3+1=2x`

`=>x^3-y^3=2y-2x`

`<=>(x-y)(x^2+xy+y^2)+2(x-y)=0`

`<=>(x-y)(x^2+xy+y^2+2)=0`

Vì `x^2+xy+y^2+2>=2>0`

`=>x-y=0<=>x=y` thay vào bthức

`=>x^3+1=2x`

`<=>x^3-2x+1=0`

`<=>x^3-x^2+x^2-2x+1=0`

`<=>x^2(x-1)+(x-1)^2=0`

`<=>(x-1)(x^2+x-1)=0`

`+)x=1=>x=y=1`

`+)x^2+x-1=0`

`\Delta=1+4=5`

`=>x_1=(-1-sqrt5)/2,x_2=(-1+sqrt5)/2`

`=>x=y=(-1-sqrt5)/2,x=y=z(-1+sqrt5)/2`

Vậy `(x,y)=(1,1),((-1-sqrt5)/2,(-1-sqrt5)/2),((-1+sqrt5)/2,(-1+sqrt5)/2)`

\(\left\{{}\begin{matrix}x^2-xy-2y^2=0\\3x+y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x\left(1-3x\right)-2\left(1-3x\right)^2=0\\y=1-3x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-14x^2+11x-2=0\\y=1-3x\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{7}\end{matrix}\right.\\y=1-3x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{2}{7}\\y=\dfrac{1}{7}\end{matrix}\right.\end{matrix}\right.\)

Vậy...

1: =>x^2+3x-4=0

=>(x+4)(x-1)=0

=>x=1 hoặc x=-4

2: =>2x-3y=1 và 3x=4y+2

=>2x-3y=1 và 3x-4y=2

=>x=2 và y=1

moi hok klop 6

\(x=7\)

ĐKXĐ:\(x\ge-9\)

\(x-\sqrt{x+9}=3\\ \Leftrightarrow\sqrt{x+9}=x-3\left(x\ge3\right)\\ \Leftrightarrow x+9=x^2-6x+9\\ \Leftrightarrow x^2-7x=0\\ \Leftrightarrow x\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy x=7