=>3/x=2/y và 96/x+1=104/y

=>2x=3y và 96/x+1=104/y

=>x/3=y/2=k và 96/x+1=104/y

=>x=3k; y=2k

\(\dfrac{96}{x}+1=\dfrac{104}{y}\)

=>\(\dfrac{96}{3k}+1=\dfrac{104}{2k}\)

=>\(\dfrac{32}{k}+1=\dfrac{52}{k}\)

=>20/k=1

=>k=20

=>x=60; y=40

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: \(\left\{{}\begin{matrix}\dfrac{x}{35}-y=2\\y-\dfrac{x}{50}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x-35y}{35}=2\\\dfrac{50y-x}{50}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-35y=70\\-x+50y=50\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15y=120\\x-35y=70\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=8\\x=70+35y=70+35\cdot8=350\end{matrix}\right.\)

b: ĐKXĐ: x<>0 và y<>0

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=\dfrac{3}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{y}=\dfrac{3}{16}-\dfrac{1}{4}=\dfrac{-1}{16}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=48\\\dfrac{1}{x}=\dfrac{1}{16}-\dfrac{1}{48}=\dfrac{2}{48}=\dfrac{1}{24}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=24\\y=48\end{matrix}\right.\left(nhận\right)\)

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)  

ĐKXĐ: \(x\ne_-^+y;y\ne0\)

Từ PT thứ 2 ta có:\(\dfrac{96}{x-y}+\dfrac{72}{x-y}=\dfrac{24}{y}\)

<=>\(\dfrac{168}{x-y}=\dfrac{24}{y}\)

<=>\(\dfrac{168}{x-y}=\dfrac{168}{7y}\)

<=>x-y=7y

<=>x=8y

Thay x=8y vào PT thứ nhất:

\(\dfrac{96}{8y+y}+\dfrac{96}{8y-y}=14\)

<=>\(\dfrac{32}{3y}+\dfrac{96}{7y}=14\)

<=>32.7y+96.3y=294y²

<=>512y=294y²

<=>y=\(\dfrac{256}{147}\left(Doy\ne0\right)\)

=>x=8y=\(\dfrac{2048}{147}\)

Vậy...

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

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

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

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

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

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