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a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
\(a)\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)\(=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{3}{10}.\frac{-4}{3}}=\frac{5}{24}\)
Hok tốt
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
a) \(\frac{\left(5.2\right)}{3.2}-\frac{1}{2}x+\frac{1}{3}+\frac{1}{5}=\frac{\left(3.2\right)}{5}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{2}x+\frac{8}{15}=\frac{6}{5}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{2}{3}=\frac{1}{2}x\)
\(\Leftrightarrow\)\(-\frac{1}{6}=\frac{1}{2}x\)
\(\Leftrightarrow\)x=-1/3
b) VT= \(\frac{\left(3.5.4.2\right)}{5.2.3}=4\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right):6+4=4:\frac{2}{3}=6\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right):6=2\)
\(\Leftrightarrow x-\frac{1}{2}=12\)
=> x= 12,5
\(\left( {\dfrac{1}{7}x - \dfrac{2}{7}} \right)\left( {\dfrac{{ - 1}}{5}x + \dfrac{3}{5}} \right)\left( {\dfrac{1}{3}x + \dfrac{1}{3}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \dfrac{1}{7}x - \dfrac{2}{7} = 0\\ \dfrac{{ - 1}}{5}x + \dfrac{3}{5} = 0\\ \dfrac{1}{3}x + \dfrac{1}{3} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \dfrac{1}{7}x = \dfrac{2}{7}\\ - \dfrac{1}{5}x = - \dfrac{3}{5}\\ \dfrac{1}{3}x = - \dfrac{1}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 3\\ x = - 1 \end{array} \right. \)