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`Answer:`
a. \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{41}{4}+\frac{3}{4}\\2x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-\frac{19}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=11:2\\x=-\frac{19}{2}:2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{19}{4}\end{cases}}\)
b. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{1}{5}\\x=-\frac{3}{5}-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
c. \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}-\left(-\frac{24}{27}\right)\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Leftrightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Leftrightarrow3x=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{9}:3\)
\(\Leftrightarrow x=\frac{1}{27}\)
a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)
\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)
hay \(x=-\dfrac{41}{12}\)
Vậy: \(x=-\dfrac{41}{12}\)
\(\frac{25}{27}:\frac{4}{7}+\frac{2}{27}:\frac{4}{7}\)
\(=\frac{4}{7}\left(\frac{25}{27}+\frac{2}{27}\right)\)
\(=\frac{4}{7}.1\)
\(=\frac{4}{7}\)
| x - 1 | = \(\frac{4}{5}\)
\(\Rightarrow\orbr{\begin{cases}\left|x-1\right|=\frac{4}{5}\\\left|x-1\right|=\frac{-4}{5}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{5}\\x=\frac{1}{5}\end{cases}}}\)
\(\frac{25}{27}:\frac{4}{7}+\frac{2}{27}:\frac{4}{7}\)
= \(\left(\frac{25}{27}+\frac{2}{27}\right):\frac{4}{7}\)
=\(1:\frac{4}{7}=\frac{7}{4}\)\(=1\frac{3}{4}\)
\(\dfrac{5}{27}-\dfrac{2}{-27}=\dfrac{5}{27}+\dfrac{2}{27}=\dfrac{7}{27}\)
\(\dfrac{7}{-25}+\dfrac{-8}{25}=\dfrac{\left(-7\right)+\left(-8\right)}{25}=\dfrac{-15}{25}=\dfrac{-3}{5}\)
\(\dfrac{23}{-11}-\dfrac{-3}{-11}=\dfrac{\left(-23\right)-3}{11}=\dfrac{-26}{11}\)
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
a) Ta có: \(\left(2x-5\right)^3=216\)
\(\Leftrightarrow2x-5=6\)
\(\Leftrightarrow2x=11\)
hay \(x=\dfrac{11}{2}\)
b) Ta có: \(2x-3⋮x+4\)
\(\Leftrightarrow-11⋮x+4\)
\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{-3;-5;7;-15\right\}\)