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a) Ta có: \(\left(x-\dfrac{3}{4}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{4}=0\)
hay \(x=\dfrac{3}{4}\)
b) Ta có: \(\left(x+\dfrac{4}{9}\right)^2=\dfrac{49}{144}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=\dfrac{7}{12}\\x+\dfrac{4}{9}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{36}\\x=\dfrac{-37}{36}\end{matrix}\right.\)
a: =>2x>-6
hay x>-3
e: =>(5-x)/x<0
=>0<x<5
h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)
\(\Leftrightarrow x+3< 0\)
hay x<-3
g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)
b/ \(\left|\left|3x-1+9\right|\right|=-\left(-31\right)\)
<=> \(\left|\left|3x+8\right|\right|=31\)
<=> \(\left|3x+8\right|=31\)
<=> \(\orbr{\begin{cases}3x+8=-31\\3x+8=31\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=-39\\3x=23\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-13\\x=\frac{23}{3}\end{cases}}\)
Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)