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=>16x+9y=840 và 210/x-210/y=7/4
=>16x=840-9y và 30/x-30/y=1/4
=>x=-9/16y+52,5 và (30y-30x)=xy/4
=>xy=120y-120x
=>y(-9/16y+52,5)=120y-120(-9/16y+52,5)
=>-9/16y^2+52,5y-120y+120(-9/16y+52,5)=0
=>-9/16y^2-67,5y-67,5y+6300=0
=>y=40 hoặc y=-280
=>x=30 hoặc x=210
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
\(\left\{{}\begin{matrix}\dfrac{4x+3y}{xy}=\dfrac{4}{11}\\\dfrac{2x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)(x,y\(\ne0\))<=>\(\left\{{}\begin{matrix}\dfrac{4}{y}+\dfrac{3}{x}=\dfrac{4}{11}\\\dfrac{2}{y}+\dfrac{1}{x}=\dfrac{4}{5}\end{matrix}\right.\)
đặt \(\dfrac{1}{x}=a\)
\(\dfrac{1}{y}=b\)
=>\(\left\{{}\begin{matrix}3a+4b=\dfrac{4}{11}\\a+2b=\dfrac{4}{5}\end{matrix}\right.< =>\left\{{}\begin{matrix}3a+4b=\dfrac{4}{11}\\3a+6b=\dfrac{12}{5}\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}-2b=-\dfrac{112}{55}\\a+2b=\dfrac{4}{5}\end{matrix}\right.< =>\left\{{}\begin{matrix}b=\dfrac{56}{55}\\a=\dfrac{-68}{55}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=a=-\dfrac{68}{55}\\\dfrac{1}{y}=b=\dfrac{56}{55}\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{-55}{68}\left(TM\right)\\y=\dfrac{55}{56}\left(TM\right)\end{matrix}\right.\)
vậy...
Thay \(x=\dfrac{3}{4}y\) vào phương trình dưới, ta có:
\(\dfrac{1}{2}\left(\dfrac{3}{4}y+3\right)\left(y-2\right)-\dfrac{1}{2}.\dfrac{3}{4}y^2=9\)
\(\Leftrightarrow\dfrac{3}{8}y^2-\dfrac{3}{4}y+\dfrac{3}{2}y-3-\dfrac{3}{8}y^2=9\\ \Leftrightarrow\dfrac{3}{4}y=12\\ \Leftrightarrow y=18\Rightarrow x=12\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(12;18\right)\)
đk x khác 7/2 ; y khác -6
Đặt \(\dfrac{1}{2x-7}=t;\dfrac{1}{y+6}=u\)
\(\left\{{}\begin{matrix}3t+4u=7\\2t-3u=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6t+8u=14\\6t-9u=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=1\\t=\dfrac{-1+3u}{2}=1\end{matrix}\right.\)
Theo cách đặt \(\left\{{}\begin{matrix}2x-7=1\\y+6=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-5\end{matrix}\right.\left(tm\right)\)
a) Ta có: \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x-y=3\\\sqrt{2}x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=1\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{4\sqrt{2}}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\2x=3+8y=3+8\cdot\dfrac{-2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
\(Đặt:a=\dfrac{1}{x};b=\dfrac{1}{y}\left(x,y\ne0\right)\\ \left\{{}\begin{matrix}\dfrac{108}{x}+\dfrac{63}{y}=7\\\dfrac{81}{x}+\dfrac{84}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}108a+63b=7\\81a+84b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}324a+189b=21\\324a+336b=28\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-147b=-7\\81a+84b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{-7}{-147}=\dfrac{1}{21}\\81a+84.\dfrac{1}{21}=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\81a=7-4=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{y}=\dfrac{1}{21}\left(TM\right)\\a=\dfrac{1}{x}=\dfrac{3}{81}=\dfrac{1}{27}\left(TM\right)\end{matrix}\right.\\ Vậy:\left\{{}\begin{matrix}x=27\\y=21\end{matrix}\right. \)
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.
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: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
Còn lại tương tự
ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{210}{x}-\dfrac{210}{y}=\dfrac{7}{4}\\4x+\dfrac{9}{4}y=210\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{30}{x}-\dfrac{30}{y}=\dfrac{1}{4}\\16x+9y=840\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}30\left(\dfrac{1}{x}-\dfrac{1}{y}\right)=\dfrac{1}{4}\\16x=840-9y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{120}\\x=\dfrac{840-9y}{16}\end{matrix}\right.\)
\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{120}\)
=>\(\dfrac{16}{840-9y}-\dfrac{1}{y}=\dfrac{1}{120}\)
=>\(\dfrac{16y-840+9y}{y\left(840-9y\right)}=\dfrac{1}{120}\)
=>\(y\left(840-9y\right)=120\left(25y-840\right)\)
=>\(-9y^2+840y-3000y+100800=0\)
=>\(-9y^2-2160y+100800=0\)
=>\(y^2+240y-11200=0\)
=>\(\left[{}\begin{matrix}y=40\left(nhận\right)\\y=-280\left(nhận\right)\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{840-9\cdot40}{16}=\dfrac{840-360}{16}=30\left(nhận\right)\\x=\dfrac{840-9\cdot\left(-280\right)}{16}=210\left(nhận\right)\end{matrix}\right.\)
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