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\(\left\{{}\begin{matrix}\dfrac{xy}{4x+3y}=\dfrac{4}{7}\\\dfrac{xy}{2x+y}=\dfrac{4}{5}\end{matrix}\right.\)\(\left(đk:4x\ne-3y,-2x\ne y,xy\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4x+3y}{xy}=\dfrac{7}{4}\\\dfrac{2x+y}{xy}=\dfrac{5}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4x+3y}{xy}=\dfrac{7}{4}\\\dfrac{4x+2y}{xy}=\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=-\dfrac{3}{4}\\\dfrac{xy}{2x+y}=\dfrac{4}{5}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\y=1\end{matrix}\right.\)
Đk: \(x\ne0,y\ne-1\)
\(\left\{{}\begin{matrix}2x+3y=xy+5\left(1\right)\\\dfrac{1}{x}+\dfrac{1}{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=xy+5\\y+1+x=x\left(y+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=xy+5\\y+1=xy\end{matrix}\right.\)
\(\Rightarrow2x+3y=y+1+5\)
\(\Leftrightarrow x=3-y\) thay vào (1) có:
\(2\left(3-y\right)+3y=\left(3-y\right)y+5\)
\(\Leftrightarrow y^2-2y+1=0\)
\(\Leftrightarrow y=1\) \(\Rightarrow x=2\)(tm)
Vậy (x;y)=(2;1)
a) \(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)(Đk: \(x\ne-1;y\ne-1\))
Đặt \(\dfrac{x}{x+1}\) là A
\(\dfrac{y}{y+1}\) là B
Ta có HPT mới : \(\left\{{}\begin{matrix}2A+B=2\\A+3B=-1\end{matrix}\right.\)(1)
Giải HPT (1) ta được A= \(\dfrac{7}{5}\) ; B=\(-\dfrac{4}{5}\)
+Với A=\(\dfrac{7}{5}\) ta có:
\(\dfrac{x}{x+1}=\dfrac{7}{5}\)
<=>\(5x=7x+7\)
<=>-2x=7
<=> x=\(-\dfrac{7}{2}\)
+Với B = \(-\dfrac{4}{5}\) ta có:
\(\dfrac{y}{y+1}=-\dfrac{4}{5}\)
<=>5y=-4y-4
<=>9y=-4
<=>y=\(-\dfrac{4}{9}\)
Vậy HPT có nghiệm (x;y) = \(\left\{-\dfrac{7}{2};-\dfrac{4}{9}\right\}\)
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
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
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix}
\frac{1}{x}+\frac{1}{y}=\frac{3}{8}\\
\frac{1}{y}+\frac{1}{z}=\frac{3}{4}\\
\frac{1}{z}+\frac{1}{x}=\frac{5}{6}\end{matrix}\right.\Rightarrow 2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{8}+\frac{3}{4}+\frac{5}{6}\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{47}{48}\)
\(\Rightarrow \left\{\begin{matrix} \frac{1}{z}=\frac{47}{48}-\frac{3}{8}\\ \frac{1}{x}=\frac{47}{48}-\frac{3}{4}\\ \frac{1}{y}=\frac{47}{48}-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{48}{29}\\ y=\frac{48}{11}\\ z=\frac{48}{7}\end{matrix}\right.\)
1.
ĐKXĐ: ....
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2x^2-1=xy\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2x-\dfrac{1}{x}=y\end{matrix}\right.\)
Trừ vế cho vế: \(\Rightarrow x=\dfrac{1}{y}\Rightarrow xy=1\)
Thay xuống pt dưới: \(2x^2-2=0\Leftrightarrow x^2=1\Leftrightarrow...\)
2.
Với \(y=0\) không phải nghiệm
Với \(y\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}4x^3+1=\dfrac{3}{y}\\3x-1=\dfrac{4}{y^3}\end{matrix}\right.\)
Cộng vế với vế:
\(4x^3+3x=4\left(\dfrac{1}{y}\right)^3+3\left(\dfrac{1}{y}\right)\)
\(\Leftrightarrow4\left(x^3-\dfrac{1}{y^3}\right)+3\left(x-\dfrac{1}{y}\right)=0\)
\(\Leftrightarrow4\left(x-\dfrac{1}{y}\right)\left(x^2+\dfrac{x}{y}+y^2\right)+3\left(x-\dfrac{1}{y}\right)=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{y}\right)\left(4x^2+\dfrac{4x}{y}+\dfrac{4}{y^2}+3\right)=0\)
\(\Leftrightarrow x-\dfrac{1}{y}=0\Leftrightarrow y=\dfrac{1}{x}\)
Thế vào pt đầu:
\(4x^3+1=3x\)
\(\Leftrightarrow4x^3-3x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)^2=0\)
\(\Leftrightarrow...\)
\(\left\{{}\begin{matrix}\dfrac{4x+3y}{xy}=\dfrac{4}{11}\\\dfrac{2x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)(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 \(\dfrac{1}{x}=a\)
\(\dfrac{1}{y}=b\)
=>\(\left\{{}\begin{matrix}3a+4b=\dfrac{4}{11}\\a+2b=\dfrac{4}{5}\end{matrix}\right.< =>\left\{{}\begin{matrix}3a+4b=\dfrac{4}{11}\\3a+6b=\dfrac{12}{5}\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}-2b=-\dfrac{112}{55}\\a+2b=\dfrac{4}{5}\end{matrix}\right.< =>\left\{{}\begin{matrix}b=\dfrac{56}{55}\\a=\dfrac{-68}{55}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=a=-\dfrac{68}{55}\\\dfrac{1}{y}=b=\dfrac{56}{55}\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{-55}{68}\left(TM\right)\\y=\dfrac{55}{56}\left(TM\right)\end{matrix}\right.\)
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