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Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
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
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: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
Chọn B