\(\left\{{}\begin{matrix}x-y=10\\\dfrac{-120\left(x-y\right)}{xy}=\dfrac{2}{5}\end{matrix}\right.\) \(\Rightarrow\dfrac{-1200}{xy}=\dfrac{2}{5}\Rightarrow xy=-3000\)

Ta được hệ: \(\left\{{}\begin{matrix}x-y=10\\xy=-3000\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y+10\\xy=-3000\end{matrix}\right.\)

Thay pt trên vào dưới:

\(\left(y+10\right).y=-3000\Rightarrow y^2+10y+3000=0\)

\(\Rightarrow\) pt vô nghiệm

Vậy hệ đã cho vô nghiệm

=>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

1) \(\dfrac{120\left(x-10\right)}{x\left(x-10\right)}-\dfrac{120x}{x\left(x-10\right)}=1\)

=> \(\dfrac{120x-1200-120x}{x\left(x-10\right)}=1\)

=> x(x-10)=-1200

=> x²-10x+1200=0

=> (x²-10x+25)+1175=0

=> (x-5)²+1175>0

=> pt vo nghiem

\(\left\{{}\begin{matrix}X+44=Y\\\dfrac{120}{X}+\dfrac{11}{30}=\dfrac{120}{Y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}X=Y-44\\3600Y+11XY=3600X\end{matrix}\right.\)

\(3600Y+11\left(Y-44\right)Y=3600\left(Y-44\right)\\ =11Y^2-484Y+158400 =0\)

\(\Delta'=\left(-242\right)^2-158400.11=-1683836\)
=> DO \(\Delta'>0\) nên pt vô nghiệm

Câu a sử dụng pp thế

Câu b thì đặt \(\left\{{}\begin{matrix}u=\dfrac{1}{x}\\v=\dfrac{1}{y}\end{matrix}\right.\) ra được hệ mới, giải tìm u,v rồi tìm x,y

\(\left\{{}\begin{matrix}\dfrac{120}{x}=\dfrac{80}{y}\\\dfrac{104}{y}-1=\dfrac{96}{x}\end{matrix}\right.\)(1)

Đặt \(a=\dfrac{1}{x}\);\(b=\dfrac{1}{y}\)

Vậy (1)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}120a=80b\\104b-1=96a\left(2\right)\end{matrix}\right.\)

Ta có \(120a=80b\Leftrightarrow b=\dfrac{3}{2}a\)

Thay \(b=\dfrac{3}{2}a\) vào (2)\(\Leftrightarrow104.\dfrac{3}{2}a-1=96a\Leftrightarrow156a-1=96a\Leftrightarrow60a=1\Leftrightarrow a=\dfrac{1}{60}\)

Vậy \(b=\dfrac{3}{2}.a=\dfrac{3}{2}.\dfrac{1}{60}=\dfrac{1}{40}\)

Vậy \(\left\{{}\begin{matrix}a=\dfrac{1}{60}\\b=\dfrac{1}{40}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=60\\y=40\end{matrix}\right.\)

Vậy (x;y)=(60;40)

\(\left\{{}\begin{matrix}\dfrac{3}{x}=\dfrac{2}{y}\\\dfrac{104}{y}-1=\dfrac{96}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{96}{x}=\dfrac{64}{y}\\\dfrac{104}{y}-1=\dfrac{96}{x}\end{matrix}\right.\) \(\Rightarrow\dfrac{104}{y}-1=\dfrac{64}{y}\)

\(\Rightarrow\dfrac{40}{y}=1\Rightarrow y=40\)

\(\Rightarrow x=\dfrac{3y}{2}=60\)

Vậy nghiệm của hệ là \(\left(x;y\right)=\left(60;40\right)\)

9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)

10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)

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

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

12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)

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

14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)

15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))

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

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

a) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{6}{y}=9\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{7}{x}=16\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{7}{16}\\y=-\dfrac{42}{17}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{7}{16};-\dfrac{42}{17}\))}

b) Đk xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{1}{y}=14\\\dfrac{8}{x}-\dfrac{1}{y}=-8\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{13}{x}=6\\\dfrac{5}{x}+\dfrac{1}{y}=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{13}{6}\\y=\dfrac{13}{152}\end{matrix}\right.\)

Vậy S={(\(\dfrac{13}{6};\dfrac{13}{152}\))}

c) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{7}{y}=21\\-\dfrac{2}{x}-\dfrac{5}{y}=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{2}{y}=10\\\dfrac{2}{x}+\dfrac{7}{y}=21\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=-\dfrac{1}{7}\end{matrix}\right.\)

Vậy S={(\(-\dfrac{1}{7};\dfrac{1}{5}\))}

d) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{9}{x}+\dfrac{2}{y}=22\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{14}{x}=35\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)

Vậy S={(0,4;-4)}

e) ĐKXĐ : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{5}{y}=10\\-\dfrac{3}{x}-\dfrac{7}{y}=8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-\dfrac{2}{y}=18\\\dfrac{3}{x}+\dfrac{5}{y}=10\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{1}{9}\\x=\dfrac{3}{55}\end{matrix}\right.\) 'Vậy....

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

Theo đề, ta có: \(\begin{cases}-3a+b=12\\ 2a-3b=-1\end{cases}\Rightarrow\begin{cases}-9a+3b=36\\ 2a-3b=-1\end{cases}\)

=>\(\begin{cases}-9a+3b+2a-3b=36-1\\ -3a+b=12\end{cases}\Rightarrow\begin{cases}-7a=35\\ b=3a+12\end{cases}\)

=>\(\begin{cases}a=-5\\ b=3\cdot\left(-5\right)+12=-15+12=-3\end{cases}\Rightarrow\begin{cases}x-y+1=-\frac15\\ x+y-2=-\frac13\end{cases}\)

=>\(\begin{cases}x-y=-\frac15-1=-\frac65\\ x+y=-\frac13+2=\frac53\end{cases}\Rightarrow\begin{cases}x=\left(-\frac65+\frac53\right):2=\left(-\frac{18}{15}+\frac{25}{15}\right):2=\frac{7}{15}:2=\frac{7}{30}\\ y=\frac53-\frac{7}{30}=\frac{50}{30}-\frac{7}{30}=\frac{43}{30}\end{cases}\)

b: \(\begin{cases}x^2+2\left(y^2+2y\right)=10\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\Rightarrow\begin{cases}3x^2+6\left(y^2+2y\right)=30\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\)

=>\(\begin{cases}3x^2+6\left(y^2+2y\right)-3x^2+\left(y^2+2y\right)=30-9\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\)

=>\(\begin{cases}7\left(y^2+2y\right)=21\\ 3x^2=\left(y^2+2y\right)+9\end{cases}\Rightarrow\begin{cases}y^2+2y=3\\ 3x^2=3+9=12\end{cases}\Rightarrow\begin{cases}y^2+2y-3=0\\ x^2=4\end{cases}\)

=>\(\begin{cases}\left(y+3\right)\left(y-1\right)=0\\ x\in\left\lbrace2;-2\right\rbrace\end{cases}\Rightarrow\begin{cases}y\in\left\lbrace-3;1\right\rbrace\\ x\in\left\lbrace2;-2\right\rbrace\end{cases}\)

c: ĐKXĐ; x>1; y>-2

\(\begin{cases}\frac{7}{\sqrt{x-1}}-\frac{5}{\sqrt{y+2}}=\frac92\\ \frac{3}{\sqrt{x-1}}+\frac{2}{\sqrt{y+2}}=4\end{cases}\Rightarrow\begin{cases}\frac{21}{\sqrt{x-1}}-\frac{15}{\sqrt{y+2}}=\frac92\cdot3=\frac{27}{2}\\ \frac{27}{\sqrt{x-1}}+\frac{18}{\sqrt{y+2}}=36\end{cases}\)

=>\(\begin{cases}\frac{21}{\sqrt{x-1}}-\frac{15}{\sqrt{y+2}}-\frac{21}{\sqrt{x-1}}-\frac{18}{\sqrt{y+2}}=\frac{27}{2}-36\\ \frac{7}{\sqrt{x-1}}-\frac{5}{\sqrt{y+2}}=\frac92\end{cases}\Rightarrow\begin{cases}-\frac{33}{\sqrt{y+2}}=\frac{27}{2}-\frac{72}{2}=\frac{-45}{2}\\ \frac{7}{\sqrt{x-1}}=\frac{5}{\sqrt{y+2}}+\frac92\end{cases}\)

=>\(\begin{cases}\sqrt{y+2}=33\cdot\frac{2}{45}=\frac{66}{45}=\frac{22}{15}\\ \frac{7}{\sqrt{x-1}}=5:\frac{22}{15}+\frac92=5\cdot\frac{15}{22}+\frac92=\frac{75}{22}+\frac{99}{22}=\frac{174}{22}=\frac{87}{11}\end{cases}\)

=>\(\begin{cases}y+2=\frac{484}{225}\\ x-1=\frac{5929}{7569}\end{cases}\Rightarrow\begin{cases}y=\frac{484}{225}-2=\frac{34}{225}\\ x=\frac{13498}{7569}\end{cases}\) (nhận)