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\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
Vậy x=\(\frac{20}{27}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)
\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)
\(\frac{9}{11}-x=\frac{-2}{11}\)
\(x=\frac{9}{11}-\frac{-2}{11}\)
\(x=1\)
Vậy x=1
\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)
\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)
\(\frac{-11}{12}\cdot x=\frac{21}{12}\)
\(x=\frac{-21}{11}\)
Vậy x=\(\frac{-21}{11}\)
\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)
\(\frac{3}{2}+x=\frac{23}{4}\)
\(x=\frac{17}{4}\)
Vậy x=\(\frac{17}{4}\)
\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)
\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)
\(\frac{3}{4}-x:\frac{2}{15}=-13\)
\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)
\(x:\frac{2}{15}=\frac{45}{4}\)
\(x=\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=1\)
\(\frac{1}{6}-x=2\)
\(x=\frac{1}{6}-2\)
\(x=\frac{-11}{6}\)
Vậy x=\(\frac{-11}{6}\)
\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)
\(1-2x=\frac{-1}{10}\)
\(2x=1-\frac{-1}{10}\)
\(2x=\frac{11}{10}\)
\(x=\frac{11}{20}\)
Vậy x=\(\frac{11}{20}\)
\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\) \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)
\(\frac{1}{2}x=\frac{11}{12}\) \(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{11}{6}\) \(x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
tk mình đi mình làm nốt cho hjhj ^^
1.
a.\(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
b. \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\)
c. \(\left(\frac{-3}{5}\right)^5=\frac{-243}{3125}\)
d. \(\left(\frac{-1}{5}\right)^2=\frac{1}{25}\)
e. \(\left(\frac{-1}{6}\right)^3=\frac{-1}{216}\)
Trả lời:
Bài 1:
a, \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{16}\)
b, \(\left(\frac{1}{2}\right)^3=\frac{1^3}{2^3}=\frac{1}{8}\)
c, \(\left(\frac{-3}{5}\right)^2=\frac{\left(-3\right)^2}{5^2}=\frac{9}{25}\)
d, \(\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)
e, \(\left(\frac{-1}{6}\right)^3=\frac{\left(-1\right)^3}{6^3}=\frac{-1}{216}\)
Bài 2:
a, \(\left(\frac{3}{2}\right)^2.\left(\frac{4}{3}\right)^2=\frac{9}{4}.\frac{16}{9}=4\)
b, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)
c, \(\left(-\frac{1}{2}\right)^2.\left(\frac{2}{5}\right)^2=\frac{1}{4}.\frac{4}{25}=\frac{1}{25}\)
d, \(\left(-\frac{1}{2}\right)^3.\left(\frac{2}{3}\right)^3=-\frac{1}{8}.\frac{8}{27}=-\frac{1}{27}\)
e, \(\left(-5\right)^3.\frac{1}{5}=-125.\frac{1}{5}=-25\)
f, \(\left(\frac{2}{9}\right)^5.\left(-\frac{27}{4}\right)^5=\frac{2^5}{9^5}.\frac{\left(-27\right)^5}{4^5}=\frac{2^5.\left(-27\right)^5}{9^5.4^5}=\frac{2^5.\left[\left(-3\right)^3\right]^5}{\left(3^2\right)^5.\left(2^2\right)^5}=-\frac{2^5.3^{15}}{3^{10}.2^{10}}=\frac{3^5}{2^5}\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
a) \(\left|x+\frac{1}{2}\right|=\frac{1}{3}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{3}\\x+\frac{1}{2}=-\frac{1}{3}\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}\)
Vậy....
b) \(\left|x-\frac{1}{2}\right|=\frac{1}{3}-\frac{1}{2}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=-\frac{1}{6}\) vô lí do \(\left|a\right|\ge0\)
Vậy pt vô nghiệm
c) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\)\(\left|x+\frac{1}{3}\right|=3\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)
Vậy..
d) \(\left|x-\frac{1}{5}\right|+\frac{1}{3}=\frac{1}{4}-\left|-\frac{3}{2}\right|\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|+\frac{1}{3}=-\frac{5}{4}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=-\frac{19}{12}\)vô lí do \(\left|a\right|\ge0\)với mọi a
Vậy pt vô nghiệm
e) \(\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left(\frac{2}{3}-\frac{1}{2}\right)\)
\(\Leftrightarrow\)\(\left|x-\frac{5}{2}\right|=\frac{7}{6}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{5}{2}=\frac{7}{6}\\x-\frac{5}{2}=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\frac{2}{3}\\x=\frac{4}{3}\end{cases}}\)
Vậy...