a: ĐKXĐ: \(\left\{{}\begin{matrix}x< >\dfrac{3}{2}y\\x< >-\dfrac{y}{3}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\\\dfrac{-5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{20}{2x-3y}+\dfrac{25}{3x+y}=-10\\-\dfrac{20}{2x-3y}+\dfrac{12}{3x+y}=84\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{37}{3x+y}=74\\-\dfrac{5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\-\dfrac{5}{2x-3y}+3:\dfrac{1}{2}=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\\dfrac{-5}{2x-3y}=15\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}11x=\dfrac{7}{6}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{66}\\3y=2x+\dfrac{1}{3}=\dfrac{7}{33}+\dfrac{1}{3}=\dfrac{6}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{2}{11}\end{matrix}\right.\)(nhận)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x< >y-2\\x< >-y+1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{7}{x-y+2}-\dfrac{5}{x+y-1}=\dfrac{9}{2}\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{14}{x-y+2}-\dfrac{10}{x+y-1}=9\\\dfrac{15}{x-y+2}+\dfrac{10}{x+y-1}=20\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{29}{x-y+2}=29\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)

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

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

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

c:

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

 \(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{3}{2x-y}-\dfrac{3}{x+y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)

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

=>x=2 và y=2x-3=4-3=1(nhận)

d:ĐKXĐ: \(\left\{{}\begin{matrix}x< >-y+1\\x< >\dfrac{1}{2}y-\dfrac{3}{2}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{19}{x+y-1}=\dfrac{19}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y-1=2\\\dfrac{15}{2}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\\dfrac{5}{2x-y+3}=7-\dfrac{15}{2}=-\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y=3\\2x-y+3=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\2x-y=-13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=-10\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=3-x=3+\dfrac{10}{3}=\dfrac{19}{3}\end{matrix}\right.\left(nhận\right)\)

e:

ĐKXĐ: \(x\ne\pm2y\)

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

=>\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{6}{x-2y}+\dfrac{8}{x+2y}=-2\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}+4:\dfrac{-6}{5}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}=-1+4\cdot\dfrac{5}{6}=-1+\dfrac{10}{3}=\dfrac{7}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{35}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{3}{70}\\2y=x-\dfrac{9}{7}=-\dfrac{87}{70}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{70}\\y=-\dfrac{87}{140}\end{matrix}\right.\left(nhận\right)\)

\(a,PT\left(1\right)\Leftrightarrow4x^2+4x+1-y^2=0\\ \Leftrightarrow\left(2x+1\right)^2-y^2=0\\ \Leftrightarrow\left(2x+y+1\right)\left(2x-y+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+y+1=0\\2x-y+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1-2x\\y=2x+1\end{matrix}\right.\)

Với \(y=-1-2x\Leftrightarrow x^2+x\left(-1-2x\right)+\left(-2x-1\right)^2=1\)

\(\Leftrightarrow x^2-x-2x^2+4x^2+4x+1=1\\ \Leftrightarrow3x^2+3x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=1\end{matrix}\right.\)

Với \(y=2x+1\Leftrightarrow x^2+x\left(2x+1\right)+\left(2x+1\right)^2=1\)

\(\Leftrightarrow x^2+2x^2+x+4x^2+4x+1=1\\ \Leftrightarrow7x^2+5x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=\dfrac{3}{7}\end{matrix}\right.\)

Vậy HPT có nghiệm \(\left(x;y\right)=\left\{\left(-1;1\right);\left(0;-1\right);\left(-\dfrac{5}{7};\dfrac{3}{7}\right)\right\}\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)-xy=100\\xy-\left(x-2\right)\left(y-2\right)=64\end{matrix}\right.\)

=>xy+3x+2y+6-xy=100 và xy-xy+2x+2y-4=64

=>3x+2y=94 và 2x+2y=68

=>x=26 và x+y=34

=>x=26 và y=8

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3+2}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5y+20-11}{y+4}=9\end{matrix}\right.\)

=>\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+1}-\dfrac{2}{y+4}=4-3=1\\\dfrac{-2}{x+1}+\dfrac{11}{y+4}=9+5-2=12\end{matrix}\right.\)

=>x+1=18/35; y+4=9/13

=>x=-17/35; y=-43/18

a) \(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4xy\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4\left(5y-5x\right)\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y=20y-20x\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y-20y+20x=0\end{matrix}\right.\)

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

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\3y-5x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-3y=xy\\5x=3y\end{matrix}\right.\)

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

b) \(\left\{{}\begin{matrix}\dfrac{1}{2x-3y}+\dfrac{5}{3x+y}=\dfrac{5}{8}\\\dfrac{2}{2x-3y}-\dfrac{5}{3x+y}=\dfrac{-3}{8}\end{matrix}\right.\)

Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)

=> hpt <=> \(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b=\dfrac{-3}{8}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b+a+5b=\dfrac{-3}{8}+\dfrac{5}{8}=0,25\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\3a=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\a=\dfrac{1}{12}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=\dfrac{1}{12}\\b=\dfrac{13}{120}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-3y}=\dfrac{1}{12}\\\dfrac{1}{3x+y}=\dfrac{13}{120}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=12\\3x+y=\dfrac{120}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{516}{143}\\y=-\dfrac{228}{143}\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}x-y=2\\y-3z=2\\-3x-2y+z=-2\end{matrix}\right.\)

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

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

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3y-5z=8\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3\left(3z+2\right)-5z=8\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-9z-6-5z=8\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-14z=14\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\left(-1\right)+2=1\\y=3\left(-1\right)+2=-1\\z=-1\end{matrix}\right.\)

Vậy...

a.

Với \(y=0\) không phải nghiệm

Với \(y\ne0\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5}{y}\\2x\left(x+y\right)+y=\dfrac{5}{y}\end{matrix}\right.\)

\(\Rightarrow3x+2=2x\left(x+y\right)+y\)

\(\Leftrightarrow2x^2+\left(2y-3\right)x+y-2=0\)

\(\Delta=\left(2y-3\right)^2-8\left(y-2\right)=\left(2y-5\right)^2\)

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

Thế vào pt đầu ...

Câu b chắc chắn đề sai

6. \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-2\end{matrix}\right.\)

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

8. \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(1-\dfrac{y}{2}\right).3\\6\left(1-\dfrac{y}{2}\right)+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(1-\dfrac{y}{2}\right)\\y=\left(VNghiệm\right)\end{matrix}\right.\Leftrightarrow\) không tồn tại x, y

(Các câu khác tương tự nhé.)

ý a ở đây bn https://hoc247.net/hoi-dap/toan-10/giai-he-pt-3x-x-2-2-y-2-va-3y-y-2-2-x-2-faq371128.html

b.

Với \(xy=0\) không là nghiệm

Với \(xy\ne0\)

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

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

\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)

\(\Leftrightarrow5-x^2=5x-2x^2\)

\(\Leftrightarrow...\)

c.

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

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

\(\Rightarrow5x^2-x\left(y+1\right)-4\left(y+1\right)^2=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(5x+4\left(y+1\right)\right)=0\)

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

Thế vào 1 trong 2 pt ban đầu...

a.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

b.

ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)

Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:

\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)

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

\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)

Thay xuống pt dưới:

\(6y+y=14\Rightarrow y=2\)

\(\Rightarrow x=4\)