Giúp tui với mn oi 45:(3x-4)=3
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\(\text{ 10-2.(4-3x)=-4 }\)
\(2\left(4-3x\right)=14\)
\(4-3x=14:2\)
\(4-3x=7\)
\(\Rightarrow3x=-3\)
\(\Rightarrow x=-1\)
\(\text{ 24:(3x-2)=-3}\)
\(3x-2=-8\)
\(\Rightarrow3x=-8+2\)
\(3x=-6\)
\(\Rightarrow x=-2\)
\(\text{-12+3(-x+7)=-18}\)
\(3\left(-x+7\right)=-6\)
\(-x+7=-6:3\)
\(-x+7=-2\)
\(\Rightarrow-x=-9\)
\(\Rightarrow x=9\)
\(\text{-45:5.(-3-2x)=3}\)
\(-9\left(-3-2x\right)=3\)
\(\Rightarrow-3-2x=\frac{-1}{3}\)
\(\Rightarrow-2x=\frac{-1}{3}+\left(-3\right)\)
\(-2x=\frac{-10}{3}\)
\(\Rightarrow x=\frac{5}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-4\right)\left(3x+1\right)< 0\)
=> TH1: \(\begin{matrix}2x-4< 0\\3x+1>0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x< 4\\3x>-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-\dfrac{1}{3}\end{matrix}\right.\) (tm)
TH2: \(\begin{matrix}2x-4>0\\3x+1< 0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x>4\\3x< -1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -\dfrac{1}{3}\end{matrix}\right.\) (vô lí)
=> \(2>x>-\dfrac{1}{3}\)
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vô 3 chấm
chọn cài đặt
chọn tự động điền và mật khẩu
ấn trình quản lí mật khẩu
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{100}\right).\)
\(=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`45 : (3x - 4) = 3`
``\(\Rightarrow3x-4=45:3\)
\(\Rightarrow3x-4=15\)
\(\Rightarrow3x=15+4\)
\(\Rightarrow3x=19\)
\(\Rightarrow x=19:3\)
\(\Rightarrow x=\dfrac{19}{3}\)
Vậy \(x=\dfrac{19}{3}\)