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(4/3-1/4.X-5/12)-2.X=8/3
(16-3.x-5)-24.X=32
16-3X-5-24X=32
11-27X=32
x=-7/9
\(\left(\frac{3}{2}-\frac{1}{4}.x-\frac{5}{12}\right)-2.x=\frac{8}{5}:\frac{3}{5}\)
\(\left(\frac{13}{12}-\frac{1}{4}.x\right)-2.x=\frac{8}{3}\)
\(\frac{13}{12}-x.\left(\frac{1}{4}-2\right)=\frac{8}{3}\)
\(x.\left(\frac{-1}{4}\right)=\frac{13}{12}-\frac{8}{3}=-\frac{19}{12}\)
\(x=\frac{-19}{12}:\frac{-1}{4}=\frac{19}{3}\)
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 \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
\(C=(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}):(\frac{5}{12}+1-\frac{7}{11})\)
\(=\left(\frac{88}{132}-\frac{33}{132}+\frac{60}{132}\right):\left(\frac{55}{132}+\frac{132}{132}-\frac{84}{132}\right)=\left(\frac{115}{132}\right):\frac{103}{132}=\frac{115}{132}.\frac{132}{103}=\frac{115}{103}\)
\(D=1\frac{1}{3}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-\frac{25}{100}.\frac{1}{2}=\frac{1}{3}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}=\frac{1}{3}+\frac{1}{2}-\frac{1}{8}=\frac{8+12-3}{24}=\frac{17}{24}\)
\(E=\left(-\frac{1}{2}\right)^2-\left(-2\right)^2-5^0=\frac{1}{4}-4-1=\frac{1-16-4}{4}=\frac{-19}{4}\)
\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)
\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)
\(\Rightarrow2x-1=\frac{-4}{63}\)
\(\Rightarrow2x=-\frac{4}{63}+1\)
\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_