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\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)
\(\Rightarrow5,5-\left|x-0,4\right|=-\frac{6}{5}\)
\(\Rightarrow-\left|x-0,4\right|=-\frac{6}{5}-5,5=-6,7\)
\(\Rightarrow\left|x-0,4\right|=6,7\)
\(\Rightarrow x-0,4=\pm6,7\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=6,7\\x-0,4=-6,7\end{cases}\Rightarrow\orbr{\begin{cases}x=7,1\\x=-6,3\end{cases}}}\)
\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)
=> \(\left|x-0,4\right|=5,5-\left[-\frac{6}{5}\right]=5,5+1,2=6,7\)
=> \(\left|x-0,4\right|=\pm6,7\)
Xét hai trường hợp :
TH1 : x - 0,4 = 6,7
=> x = 6,7 + 0,4 = 7,1
TH2 : x - 0,4 = -6,7
=> x = -6,7 + 0,4 =-6,3
\(b,\left[1-\frac{3}{4}\left|x\right|\right]^2=\frac{16}{25}\)
=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\sqrt{\frac{16}{25}}\)
=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\frac{4}{5}\)
=> \(\orbr{\begin{cases}1-\frac{3}{4}\left|x\right|=\frac{4}{5}\\1-\frac{3}{4}\left|x\right|=-\frac{4}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\pm\frac{4}{15}\\x=\pm\frac{12}{5}\end{cases}}\)
\(c,\left[0,1\left|x\right|-\frac{1}{2}\right]\left[0,5-\left|x\right|\right]=0\)
=> \(\orbr{\begin{cases}0,1\left|x\right|-\frac{1}{2}=0\\0,5-\left|x\right|=0\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{1}{10}\left|x\right|=\frac{1}{2}\\\left|x\right|=0,5\end{cases}}\)
=> \(\orbr{\begin{cases}\left|x\right|=5\\\left|x\right|=0,5\end{cases}}\)=> \(\orbr{\begin{cases}x\in\left\{5;-5\right\}\\x\in\left\{0,5;-0,5\right\}\end{cases}}\)
d, Xét hai trường hợp rồi ra kết quả thôi
1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
\(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
\(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
\(2x=\frac{53}{30}\)
\(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
\(2x=\frac{37}{30}\)
\(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
\(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
\(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
\(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
\(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
\(-\frac{5}{7}x=-\frac{11}{45}\)
\(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}
a: =>x-8/5=1/20-1/10=-1/20
=>x=-0,05+1,6=1,55
b: =>x-3/2=4/3 hoặc x-3/2=-4/3
=>x=17/6 hoặc x=1/6
c: =>\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{35}{12}\)
=>x-1/3=35/12 hoặc x-1/3=-35/12
=>x=39/12=13/4 hoặc x=-31/12
d: =>|x-5/8|=3/4
=>x-5/8=3/4 hoặc x-5/8=-3/4
=>x=11/8 hoặc x=-1/8
\(|x+1,5|=2\)
\(|x+\frac{3}{2}|=2\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{2}=2\\x+\frac{3}{2}=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2-\frac{3}{2}\\x=-2-\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)
Vậy \(x=\left[{}\begin{matrix}\frac{1}{2}\\\frac{-7}{2}\end{matrix}\right.\)