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Tìm x, biết:
\(\left(\frac{2}{3}\cdot x-\frac{1}{5}\right)^2=\frac{4}{9}\)
\(\Rightarrow\left(\frac{2}{3}\cdot x-\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\left(-\frac{2}{3}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}\cdot x-\frac{1}{5}=\frac{2}{3}\\\frac{2}{3}\cdot x-\frac{1}{5}=-\frac{2}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}\cdot x=\frac{13}{15}\\\frac{2}{3}\cdot x=-\frac{7}{15}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{13}{10}\\x=-\frac{7}{10}\end{cases}}\)
a)\(\frac{9}{4}\cdot\left|x\right|-\frac{5}{2}=\frac{8}{3}\)\(\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{8}{3}+\frac{5}{2}\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{31}{6}\)
\(\Rightarrow\left|x\right|=\frac{31}{6}:\frac{9}{4}\Rightarrow\left|x\right|=\frac{31}{6}\cdot\frac{4}{9}\Rightarrow\left|x\right|=\frac{62}{27}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{62}{27}\\x=-\frac{62}{27}\end{cases}}\)
b)\(\frac{1}{2}\cdot\left|x\right|+\frac{3}{4}=\frac{2}{3}\Rightarrow\frac{1}{2}\cdot\left|x\right|=\frac{2}{3}-\frac{3}{4}\Rightarrow\frac{1}{2}\cdot\left|x\right|=-\frac{1}{12}\)
\(\Rightarrow\left|x\right|=-\frac{1}{12}:\frac{1}{2}\Rightarrow\left|x\right|=-\frac{1}{12}\cdot2\Rightarrow\left|x\right|=-\frac{1}{6}\)
Ta có\(\left|x\right|\ge0\)mà \(-\frac{1}{6}\le0\)
Do đó ko có giá trị của x thỏa mãn
\(\left(3-\frac{9}{10}-\left|x+2\right|\right):\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=1\)
\(\left(\frac{21}{10}-\left|x+2\right|\right):\frac{1}{2}+\frac{4}{5}=1\)
\(\left(\frac{21}{10}-\left|x+2\right|\right):\frac{1}{2}=\frac{1}{5}\)
\(\frac{21}{10}-\left|x+2\right|=\frac{1}{10}\)
=> |x+2| = 2
TH1: x + 2 = 2 => x = 0
TH2: x + 2 = -2 => x = -4
KL:...
2(x-1)-5(x+2)=-9
2x-2 -5x-10 = -9
2x-5x = -9 +2 +10
-3x = 3
x=-1