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\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{9}{4}-2x=0\\3x+\dfrac{7}{8}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{9}{4}\\3x=-\dfrac{7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{7}{24}\end{matrix}\right.\)
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\(\left|3x-4\right|+5-2x=0\)
\(\Leftrightarrow\left|3x-4\right|=2x-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5\ge0\\\left[{}\begin{matrix}3x-4=2x-5\\3x-4=5-2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\\left[{}\begin{matrix}x=-1\left(l\right)\\x=\dfrac{9}{5}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(S=\)
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\(\left|3x-4\right|+\left|3y+5\right|=0\)
\(\Rightarrow\hept{\begin{cases}3x-4=0\\3y+5=0\end{cases}\Rightarrow\hept{\begin{cases}3x=4\\3y=-5\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=-\frac{5}{3}\end{cases}}}}\)
Vậy x=4/3 và y= -5/3
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a: 2x-1=0
nên 2x=1
hay x=1/2
b: 4x2-16=0
=>(x-2)(x+2)=0
=>x=2 hoặc x=-2
c: x2-2x=0
=>x(x-2)=0
=>x=0 hoặc x=2
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Ta có:\(\left(2x-\dfrac{1}{2}\right)\left(\dfrac{5}{4}-3x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=0\\\dfrac{5}{4}-3x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}\\3x=\dfrac{5}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{5}{12}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{5}{12}\end{matrix}\right.\)
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Vì \(\left|3x-4\right|\ge0;\left|5y+5\right|\ge0\)
\(\Rightarrow\left|3x-4\right|+\left|5y+5\right|\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|3x-4\right|=0\\\left|5y+5\right|=0\end{cases}\Rightarrow\hept{\begin{cases}3x=4\\5y=-5\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{4}{3}\\y=-1\end{cases}}}\)
5x=4
x=4/5
Ta có: \(x^2+3x-4=0\)
\(\Rightarrow x^2+3x=4\)
\(\Rightarrow x\left(x+3\right)=4\)
.................... Đến đây dễ rồi.