Câu 14:
a: \(\dfrac{2}{3}x+\dfrac{2}{5}=\dfrac{3}{10}\)
=>\(\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{2}{5}=-\dfrac{1}{10}\)
=>\(x=-\dfrac{1}{10}:\dfrac{2}{3}=\dfrac{-1}{10}\cdot\dfrac{3}{2}=\dfrac{-3}{20}\)
b: \(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)
=>\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{27}{12}=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c: \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}+x\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{4}{9}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Câu 13:
a: \(A=\dfrac{3}{7}\cdot\dfrac{2}{11}+\dfrac{3}{7}\cdot\dfrac{9}{11}\)
\(=\dfrac{3}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)\)
\(=\dfrac{3}{7}\cdot\dfrac{11}{11}=\dfrac{3}{7}\)
b: \(B=\dfrac{6}{5}\cdot\sqrt{\dfrac{25}{16}}-\left(\dfrac{3}{4}\right)^2:0,25\)
\(=\dfrac{6}{5}\cdot\dfrac{5}{4}-\dfrac{9}{16}:\dfrac{1}{4}\)
\(=\dfrac{6}{4}-\dfrac{9}{4}=-\dfrac{3}{4}\)
c: \(C=\left|\dfrac{2}{3}-1\right|-\dfrac{5}{2}\cdot\sqrt{\dfrac{4}{25}}\)
\(=\left|-\dfrac{1}{3}\right|-\dfrac{5}{2}\cdot\dfrac{2}{5}\)
\(=\dfrac{1}{3}-1=-\dfrac{2}{3}\)
