Ta có: \(\begin{cases} xy(4xy + y + 4) = y^2(2y + 5) - 1 & \text{(1)} \\ 2xy(x - 2y) + x - 14y = 0 & \text{(2)} \end{cases}\)

Khi y=0 thì (1) sẽ trở thành: \(x\cdot0\left(4\cdot x\cdot0+0+4\right)=0^2\left(2\cdot0+5\right)-1\)

=>-1=0(vô lý)

=>y<>0

Đặt u=xy

=>\(x=\frac{u}{y}\)

(2)=> \(2x^2y - 4xy^2 + x - 14y = 0\)

=>\(2u^2 + u - 4uy^3 - 14y^2 = 0\)

=>\(2u^2+(1-4y^3)u-14y^2=0\quad\) (*)

=>\(4u^2+\left(2-8y^3\right)u-28y^2=0\) (3)

(1) =>\(4x^2y^2 + xy^2 + 4xy = 2y^3 + 5y^2 - 1\)

=>\(4u^2+(y^2+4)u-(2y^3+5y^2-1)=0\left(4\right)\)

Từ (3),(4) suy ra \([(y^2 + 4) - 2(1 - 4y^3)]u - (2y^3 + 5y^2 - 1) + 28y^2 = 0\)

=>\((8y^3 + y^2 + 2)u + (-2y^3 + 23y^2 + 1) = 0\)

=>\(u = \frac{2y^3 - 23y^2 - 1}{8y^3 + y^2 + 2}\)

Thay u vào ( *), ta được:

\(y^2(y - 1)\left(-64y^6 - 232y^5 - 364y^4 - 338y^3 - 964y^2 + 15y + 11\right) = 0\)

Nếu y=1 thì \(u = \frac{2(1)^3 - 23(1)^2 - 1}{8(1)^3 + 1^2 + 2} = \frac{-22}{11} = -2\)

xy=-2

=>x*1=-2

=>x=-2

Thay x=-2; y=1 vào (1), ta được:

\((-2)\cdot1\cdot\left\lbrack4\cdot(-2)\cdot(1)+1+4\right\rbrack=1^2\left\lbrack2\cdot(1)+5\right\rbrack-1\iff6=6\) (đúng)

Thay x=-2; y=1 vào (2), ta được:

\(2\cdot(-2)\cdot1\cdot(-2-2\cdot1)+(-2)-14\cdot1=0\iff16-16=0\) (đúng)

Vậy: x=-2; y=1

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

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

Vậy...

`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)`

moi hok klop 6

\(\hept{\begin{cases}x+2y=3\\-2x-y=6\end{cases}< =>\hept{\begin{cases}x=3-2y\\-2\left(3-2y\right)-y\end{cases}< =>\hept{\begin{cases}x=3-2y\\-6+4y=6\end{cases}< =>\hept{\begin{cases}x=3-2y\\4y=12\end{cases}< =>\hept{\begin{cases}x=-3\\y=3\end{cases}}}}}}\)

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)