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a) Ta có: \(\left(-12\right)-\left|13-x\right|=-21\)
\(\Leftrightarrow-\left|x-13\right|-12=-21\)
\(\Leftrightarrow-\left|x-13\right|=-9\)
\(\Leftrightarrow\left|x-13\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=9\\x-13=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=22\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{22;4\right\}\)
b) Ta có: \(8\le\left|x\right|< 9\)
\(\Leftrightarrow\left|x\right|=8\)
\(\Leftrightarrow x\in\left\{8;-8\right\}\)
Vậy: \(x\in\left\{8;-8\right\}\)
c) Ta có: \(x-\left(-25+x\right)=13-x\)
\(\Leftrightarrow x+25-x-13+x=0\)
\(\Leftrightarrow x+12=0\)
hay x=-12
Vậy: x=-12
d) Ta có: \(\left(15-30\right)+x=x-\left(27-\left|-8\right|\right)\)
\(\Leftrightarrow x-15=x-\left(27-8\right)\)
\(\Leftrightarrow x-15-x+19=0\)
\(\Leftrightarrow-4=0\)(vô lý)
Vậy: \(x\in\varnothing\)
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
Bài 1:
a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)
