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Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
Bài 2:
1.Thay m=3, ta có:
\(\left\{{}\begin{matrix}3x+2y=5\\2x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Bài 1:
\(\left\{{}\begin{matrix}\left|x+1\right|+\left|y-1\right|=5\\\left|x+1\right|-4y=-4\end{matrix}\right.\)
\(\Rightarrow\left|y-1\right|-4y=9\)\(\Leftrightarrow\left[{}\begin{matrix}y=-3,\left(3\right)\left(KTM\right)\left(ĐK:y\ge1\right)\\y=-1,6\left(TM\right)\left(ĐK:y< 1\right)\end{matrix}\right.\)
Thay y=-1,6 vào hpt, ta được:
\(\left\{{}\begin{matrix}\left|x+1\right|=2,4\\\left|x+1\right|=-10,4\left(vl\right)\end{matrix}\right.\)
Vậy pt vô nghiệm.
1.
a, \(\left\{{}\begin{matrix}2x-3y=3\\-4x=3x-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=3\\-4x-3x=13\end{matrix}\right.\)\(\left\{{}\begin{matrix}-4x+6y=-6\\-4x-3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9y=-19\\-4x+6y=-6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=-\dfrac{19}{9}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=3\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=9\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=2\\\dfrac{3}{x}+\dfrac{3}{y}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\left(TM\right)\\y=\dfrac{1}{2}\left(TM\right)\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{5}{y}=1\\\dfrac{2}{x}+\dfrac{1}{y}=3\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{5}{y}=1\\\dfrac{10}{x}+\dfrac{5}{y}=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{x}=16\\\dfrac{10}{x}+\dfrac{5}{y}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{16}\left(TM\right)\\y=\dfrac{13}{7}\left(TM\right)\end{matrix}\right.\)
d, \(\left\{{}\begin{matrix}\sqrt{x+1}-3\sqrt{y-1}=-4\\2\sqrt{x+1}-\sqrt{y-1}=2\end{matrix}\right.\left(x\ge-1,y\ge1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+1}-6\sqrt{y-1}=-8\\2\sqrt{x+1}-\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-5\sqrt{y-1}=-10\\2\sqrt{x+1}-6\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y-1}=2\\2\sqrt{x+1}-6\sqrt{y-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\y=5\left(TM\right)\end{matrix}\right.\)
1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\)
3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
1. \(\Leftrightarrow\left\{{}\begin{matrix}mx+m^2y=3m\\mx+4y=6\end{matrix}\right.\)
\(\Rightarrow\left(m^2-4\right)y=3\left(m-2\right)\)
\(\Leftrightarrow\left(m-2\right)\left(m+2\right)y=3\left(m-2\right)\)
Để pt có nghiệm duy nhất \(\Rightarrow\left(m-2\right)\left(m+2\right)\ne0\Rightarrow m\ne\pm2\)
Để pt vô nghiệm \(\Rightarrow\left\{{}\begin{matrix}\left(m-2\right)\left(m+2\right)=0\\3\left(m-2\right)\ne0\end{matrix}\right.\) \(\Rightarrow m=-2\)
2. Không thấy m nào ở hệ?
3. Bạn tự giải câu a
b/ \(\left\{{}\begin{matrix}6x+2my=2m\\\left(m^2-m\right)x+2my=m^2-m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=\frac{\left(m-1\right)\left(1-x\right)}{2}\\\left(m^2-m-6\right)x=m^2-3m\end{matrix}\right.\)
Để hệ có nghiệm duy nhất \(\Rightarrow m^2-m-6\ne0\Rightarrow m\ne\left\{-2;3\right\}\)
Khi đó: \(\left\{{}\begin{matrix}x=\frac{m^2-3m}{m^2-m-6}=\frac{m}{m+2}\\y=\frac{\left(m-1\right)\left(1-x\right)}{2}=\frac{m-1}{m+2}\end{matrix}\right.\)
\(x+y^2=1\Leftrightarrow\frac{m}{m+2}+\frac{\left(m-1\right)^2}{\left(m+2\right)^2}=1\)
\(\Leftrightarrow m\left(m+2\right)+\left(m-1\right)^2=\left(m+2\right)^2\)
\(\Leftrightarrow m^2-4m-3=0\Rightarrow\) bấm máy, số xấu
4.
\(\Leftrightarrow\left\{{}\begin{matrix}m^2x+my=2m^2\\x+my=m+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-1\right)x=2m^2-m-1=\left(2m+1\right)\left(m-1\right)\\y=2m-mx\end{matrix}\right.\)
- Với \(m=1\) hệ có vô số nghiệm
- Với \(m=-1\) hệ vô nghiệm
- Với \(m\ne\pm1\) hệ có nghiệm duy nhất:
\(\left\{{}\begin{matrix}x=\frac{\left(2m+1\right)\left(m-1\right)}{\left(m-1\right)\left(m+1\right)}=\frac{2m+1}{m+1}\\y=2m-mx=\frac{m}{m+1}\end{matrix}\right.\)
Lời giải:
HPT \( \Leftrightarrow \left\{\begin{matrix} \sqrt{2}y=6-(m+2018)x\\ 4x+(m+2018).\sqrt{2}y=9\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow 4x+(m+2018)[6-(m+2018)x]=9\sqrt{2}\)
\(\Leftrightarrow x[4-(m+2018)^2]=9\sqrt{2}-6(m+2018)\)
\(\Leftrightarrow -x(m+2020)(m+2016)=9\sqrt{2}-6(m+2018)(*)\)
Để HPT ban đầu có nghiệm duy nhất thì PT $(*)$ phải có nghiệm $x$ duy nhất
Điều này xảy ra khi $(m+2020)(m+2016)\neq 0$
$\Leftrightarrow m\neq -2020$ và $m\neq -2016$
Lời giải:
HPT \( \Leftrightarrow \left\{\begin{matrix} \sqrt{2}y=6-(m+2018)x\\ 4x+(m+2018).\sqrt{2}y=9\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow 4x+(m+2018)[6-(m+2018)x]=9\sqrt{2}\)
\(\Leftrightarrow x[4-(m+2018)^2]=9\sqrt{2}-6(m+2018)\)
\(\Leftrightarrow -x(m+2020)(m+2016)=9\sqrt{2}-6(m+2018)(*)\)
Để HPT ban đầu có nghiệm duy nhất thì PT $(*)$ phải có nghiệm $x$ duy nhất
Điều này xảy ra khi $(m+2020)(m+2016)\neq 0$
$\Leftrightarrow m\neq -2020$ và $m\neq -2016$