ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+y}+6x-3y=6\\\dfrac{3}{x+y}+2x-4y=1\end{matrix}\right.\)

\(\Rightarrow4x+y=5\Rightarrow y=5-4x\)

Thế vào phương trình đầu:

\(\dfrac{1}{x+5-4x}+2x-\left(5-4x\right)=2\)

\(\Leftrightarrow\dfrac{1}{5-3x}+6x-7=0\)

\(\Leftrightarrow\left(6x-7\right)\left(5-3x\right)+1=0\)

\(\Leftrightarrow...\)

a: =>xy-2x+2y-4=xy+y và 5xy+10x+y+2=5xy-10x-2y+4

=>-2x+y=4 và 20x+3y=2

=>x=-5/13; y=42/13

b: =>4x+2|y|=8 và 4x-3y=1

=>2|y|-3y=7 và 4x-3y=1

TH1: y>=0

=>2y-3y=7 và 4x-3y=1

=>-y=7 và 4x-3y=1

=>y=-7(loại)

TH2: y<0

=>-2y-3y=7 và 4x-3y=1

=>y=-7/5; 4x=1+3y=1-21/5=-16/5

=>x=-4/5; y=-7/5

Đk: \(y\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=1\\\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)=-1\end{matrix}\right.\)

\(\Rightarrow-\left(x+\dfrac{1}{y}\right)^2+\dfrac{x}{y}=\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)\)

\(\Leftrightarrow-\left(x+\dfrac{1}{y}\right)^2+2\left(x+\dfrac{1}{y}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{y}=0\\x+\dfrac{1}{y}=2\end{matrix}\right.\)

TH1: \(x+\dfrac{1}{y}=0\Leftrightarrow\dfrac{1}{y}=-x\) thay vào pt dưới ta được:

\(-x^2=-1\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\Rightarrow y=-1\\x=-1\Rightarrow y=1\end{matrix}\right.\)

TH2: \(x+\dfrac{1}{y}=2\Leftrightarrow\dfrac{1}{y}=2-x\) thay vào pt dưới ta được:

\(\left(2-x\right)x-2.2=-1\)\(\Leftrightarrow x^2-2x+3=0\left(vn\right)\) 

Vậy (x;y)=(-1;1);(1;-1)

gợi ý \(\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=1\left(1\right)\\\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)=-1\left(2\right)\end{matrix}\right.\)

Đem \(\left(1\right)+\left(2\right):\left(x+\dfrac{1}{y}\right)^2-2\left(x+\dfrac{1}{y}\right)=0\)

đến đây chắc bạn có thể tự làm được

a.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)

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

b.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}x+y=-1\\\dfrac{1}{x}-\dfrac{2}{y}=2\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\\dfrac{-y-2\left(y+1\right)}{y\left(y+1\right)}=\dfrac{2y^2+2y}{y\left(y+1\right)}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1-y\\-y-2y-2=2y^2+2y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-y-1\\2y^2+5y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y-1\\\left(2y+1\right)\left(y+2\right)=0\end{matrix}\right.\)

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

Vậy: \(\left(x,y\right)\in\left\{\left(-\dfrac{1}{2};-\dfrac{1}{2}\right);\left(1;-2\right)\right\}\)

ĐK : x ≠ 0 ; y ≠ 0

\(\left\{{}\begin{matrix}x+y=-1\\\dfrac{1}{x}-\dfrac{2}{y}=2\end{matrix}\right.\text{⇔}\left\{{}\begin{matrix}x=-1-y\\\dfrac{1}{-1-y}-\dfrac{2}{y}=2\end{matrix}\right.\)\(\text{⇔}\left\{{}\begin{matrix}x=-1-y\\y-2\left(-1-y\right)=2y\left(-1-y\right)\end{matrix}\right.\text{⇔}\left\{{}\begin{matrix}x=-1-y\\y+2+2y=-2y-2y^2\end{matrix}\right.\)

\(\text{⇔}\left\{{}\begin{matrix}x=-1-y\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=-2\end{matrix}\right.\end{matrix}\right.\)

Với \(y=-\dfrac{1}{2}\text{⇔}x=-\dfrac{1}{2}\)

Với \(y=-2\text{⇔}x=1\)

Vậy.....

Ta có: x+y=-1

nên x=-1-y

Thay x=-1-y vào \(\dfrac{1}{x}-\dfrac{2}{y}=2\), ta được:

\(\dfrac{1}{-y-1}-\dfrac{2}{y}=2\)

\(\Leftrightarrow\dfrac{-1}{y+1}-\dfrac{2}{y}=2\)

\(\Leftrightarrow\dfrac{-y-2\left(y+1\right)}{y\left(y+1\right)}=\dfrac{2y\left(y+1\right)}{y\left(y+1\right)}\)

Suy ra: \(2y^2+2y+y+2y+2=0\)

\(\Leftrightarrow2y^2+4y+y+2=0\)

\(\Leftrightarrow\left(y+2\right)\left(2y+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\\2y+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-\dfrac{1}{2}\end{matrix}\right.\)

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

Vậy: \(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-\dfrac{1}{2};-\dfrac{1}{2}\right)\right\}\)

Lời giải:
$x+y=-1\Leftrightarrow x=-1-y$. Thay vô PT $(2)$:

$\frac{1]{-1-y}-\frac{2}{y}=2$

$\Leftrightarrow \frac{1}{y+1}+\frac{2}{y}=-2$

$\Leftrightarrow \frac{3y+2}{y(y+1)}=-2$ ($y\neq 0; -1$)

$\Rightarrow 3y+2=-2y(y+1)$

$\Leftrightarrow 3y+2+2y^2+2y=0$

$\Leftrightarrow 2y^2+5y+2=0$

$\Leftrightarrow (2y+1)(y+2)=0$

$\Rightarrow y=\frac{-1}{2}$ hoặc $y=-2$

$\Leftrightarrow x=\frac{-1}{2}$ hoặc $x=1$ (tương ứng)

ĐKXĐ: \(x;y\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+\left(y-1\right)^2=85\\x=3y\end{matrix}\right.\)

Thế pt dưới lên trên:

\(\left(3y\right)^2+\left(y-1\right)^2=85\)

\(\Leftrightarrow10y^2-2y-84=0\)

\(\Rightarrow\left[{}\begin{matrix}y=3\Rightarrow x=9\\y=-\dfrac{14}{5}\Rightarrow x=-\dfrac{42}{5}\end{matrix}\right.\)

ĐK: \(x,y\neq 0\)

\(PT\left(2\right)\Leftrightarrow x=9-y\)

Thay vào \(PT\left(1\right)\Leftrightarrow\dfrac{1}{9-y}+\dfrac{1}{y}=\dfrac{1}{2}\Leftrightarrow2y+18-2y=9y-y^2\)

\(\Leftrightarrow y^2-9y+18=0\\ \Leftrightarrow\left[{}\begin{matrix}y=3\Rightarrow x=6\\y=6\Rightarrow x=3\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(6;3\right);\left(3;6\right)\)

ĐK x;y \(\ne\)0 

HPT <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\\x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{1}{2}\\x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=18\\x+y=9\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x\left(9-x\right)=18\\y=9-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9x+18=0\\y=9-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x-6\right)=0\\y=9-x\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\\y=9-x\end{matrix}\right.\)

Khi x=  3 => y = 6

Khi x = 6 => y = 3

Vậy nghiệm phương trình là (3;6) ; (6;3)