ĐKXĐ: \(xy\ne0;x\ne\pm y\)

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

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a+b+\dfrac{1}{a+b}=\dfrac{5}{2}\\b-a+\dfrac{1}{b-a}=\dfrac{10}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-\dfrac{5}{2}\left(a+b\right)+1=0\\\left(b-a\right)^2-\dfrac{10}{3}\left(b-a\right)+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=2\\a+b=\dfrac{1}{2}\end{matrix}\right.\\\left[{}\begin{matrix}b-a=3\\b-a=\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a+b=2\\b-a=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=\dfrac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{5}{2}\end{matrix}\right.\)

3 TH còn lại xét tương tự

\(\left\{{}\begin{matrix}\left(x+10\right)\left(y-\dfrac{1}{2}\right)=xy\\\left(x-10\right)\left(y+\dfrac{1}{3}\right)=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy-\dfrac{1}{2}x+10y-5=xy\\xy+\dfrac{1}{3}x-10y-\dfrac{10}{3}=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{1}{2}x+10y=5\\\dfrac{1}{3}x-10y=\dfrac{10}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{1}{6}x=5+\dfrac{10}{3}=\dfrac{25}{3}\\-\dfrac{1}{2}x+10y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{25}{3}\cdot6=-50\\10y=5+\dfrac{1}{2}x=5+\dfrac{1}{2}\cdot\left(-50\right)=-20\end{matrix}\right.\)

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

a: \(\left\{{}\begin{matrix}\dfrac{-5x+2y}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4\left(-5x+2y\right)+60=3\left(y+27\right)-24x\\7\left(x+1\right)+21y=3\left(6y-5x\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-20x+8y+60=3y+81-24x\\7x+7+21y=18y-15x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-20x+8y-3y+24x=21\\7x+21y-18y+15x=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+5y=21\\22x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+15y=63\\110x+15y=-35\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-98x=98\\4x+5y=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\5y=21-4x=21+4=25\end{matrix}\right.\)

=>x=-1 và y=5

b: \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(xy+3x+2y+6\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(xy-2x-2y+4\right)=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+3x+2y+6-xy=100\\xy-\left(xy-2x-2y+4\right)=64\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x+2y=94\\2x+2y=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=34\\2x+2y=60\end{matrix}\right.\)

=>x=34 và y=-4

c: \(\left\{{}\begin{matrix}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{matrix}\right.\)

\(\left\{{}\begin{matrix}xy-x+20y-20=xy\\xy+x-10y-10=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x+20y=20\\x-10y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10y=30\\x-10y=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x=10y+10=30+10=40\end{matrix}\right.\)

d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-2y\\x< >-\dfrac{y}{2}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}\dfrac{4}{x+2y}+\dfrac{2}{2x+y}=6\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{2x+y}=5\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=1-2y=1-\dfrac{2}{3}=\dfrac{1}{3}\end{matrix}\right.\)(nhận)

e: ĐKXĐ: x<>-1 và y<>-4

\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4\\2-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=\dfrac{27}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{9}\\y=-\dfrac{87}{19}\end{matrix}\right.\left(nhận\right)\)

(P/s: Đang cần gấp!!!)

a)\(\left\{{}\begin{matrix}\left(x+3\right)\left(y-5\right)=xy\\\left(x-2\right)\left(y+5\right)=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy-5x+3y-15=xy\\xy+5x-2y-10=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x+3y-15=0\\5x+2y-10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=15\left(1\right)\\5x+2y=10\left(2\right)\end{matrix}\right.\)\(\left(1\right)-\left(2\right)=-y=-25\Leftrightarrow y=25\)

thay y = 25 vào \(\left(2\right)\), ta có: \(5x-2.25=10\Leftrightarrow x=12\)

Vậy hệ phương trình có nghiệm (x; y) là (12; 25)

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\dfrac{1}{6x}+\dfrac{1}{5y}=\dfrac{2}{15}\end{matrix}\right.\)

pt 2 nhân cho 30 ta đc hệ mới là

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\dfrac{5}{x}+\dfrac{6}{y}=4\end{matrix}\right.\)

đến đây ta chỉ việc đặt 1/x=a; 1/y =b

phần còn lại tương đối dễ dàng để bạn thử sức

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}\left(xy+3x+2y+6\right)=\dfrac{1}{2}xy+56\\\dfrac{1}{2}\left(xy-2x-2y+4\right)=\dfrac{1}{2}xy-32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y+6=112\\-2x-2y+4=-64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=106\\-2x-2y=-68\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=106\\x=38\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=38\\y=-4\end{matrix}\right.\)