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![](https://rs.olm.vn/images/avt/0.png?1311)
\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)
\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)
\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)
Đến đây tìm được rồi nhé
3,4, áp dụng bài 1,2 rồi làm :v
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{x}{3}\)\(-\)\(\frac{1}{8}\)\(=\)\(\frac{5}{8}\)
\(\Rightarrow\) \(\frac{x}{3}\)=\(\frac{5}{8}+\frac{1}{8}\)
\(\Rightarrow\) \(\frac{x}{3}\)\(=\frac{3}{4}\)
\(\Rightarrow\) x=3.3:4
\(\Rightarrow\) x=2,25
Vậy x=2,25
b) \(3\frac{1}{3}\).x-\(6\frac{3}{4}=3\frac{1}{4}\)
\(\Rightarrow\)\(\frac{10}{3}\).x-\(\frac{27}{4}=\frac{13}{4}\)
\(\Rightarrow\) \(\frac{10}{3}\).x=\(\frac{13}{4}+\frac{27}{4}\)
\(\Rightarrow\) \(\frac{10}{3}\).x=10
\(\Rightarrow\) x=\(10\div\frac{10}{3}\)
\(\Rightarrow\) x=3
Vậy x=3
Ủng hộ mik nha mọi người !
Ta có : \(\frac{x}{3}-\frac{1}{8}=\frac{5}{8}\)
\(\Leftrightarrow\frac{x}{3}=\frac{5}{8}-\frac{1}{8}\)
\(\Leftrightarrow\frac{x}{3}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}.\frac{1}{3}=\frac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{10}{3}x+\frac{67}{4}=-\frac{53}{4}\)
<=> \(\frac{10}{3}x=-30\)
=> x = -9
![](https://rs.olm.vn/images/avt/0.png?1311)
(4,53 x 02+6,165 x 3):2,5
=(0,906 + 18,495); 2,5
=19,401: 2,5
=7,7604
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+\frac{1}{3}\right)^4\ge0\Rightarrow A\ge-1\)
Dấu \(=\)xảy ra khi \(2x+\frac{1}{3}=0\Leftrightarrow x=-\frac{1}{6}\).
b) \(\left(\frac{4}{9}x-\frac{2}{15}\right)^6\ge0\Rightarrow B\le3\)
Dấu \(=\)xảy ra khi \(\frac{4}{9}x-\frac{2}{15}=0\Leftrightarrow x=\frac{3}{10}\).
\(\frac{3}{x-5}=-\frac{4}{x+2}\)
\(\Rightarrow3x+6=-4x+20\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=2\)
Vậy: \(x=2\)
\(\frac{x}{-2}=-\frac{8}{x}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
Vậy: \(x\in\left\{4;-4\right\}\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Leftrightarrow3\left(x-2\right)=-4\left(x-5\right)\)
\(3x-6=-4x+20\)
....