Giải hộ e với: 9^x-1=1/9
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^{x-1}=\frac{1}{9}\)
\(9^{x-1}=9^{-1}\)
=> x - 1 = - 1
x = -1
x = 0
9x-1= \(\frac{1}{9}\)
9x-1= 9-1
=> x-1=-1
x= -1-1
x= 0
tk nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
<=> \(\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
<=> \(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
<=> \(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{\frac{x-1}{x+1}}=\frac{2}{9}\right)\)
<=> \(2.\left(\frac{1}{6}-\frac{\frac{1}{x-1}}{x+1}\right)=\frac{2}{9}\)
<=> \(\frac{1}{6}+\frac{1}{x+1}=\frac{1}{9}\)
<=> \(\frac{1}{x-1}=\frac{1}{6}-\frac{1}{9}\)
<=> \(\frac{1}{x+1}=\frac{1}{18}\)
<=> x + 1 = 18
=> x = 18 - 1
x = 17
![](https://rs.olm.vn/images/avt/0.png?1311)
(x-1)^2=9
=> (x-1)^2 = 3^2
=> x-1= 3
=> x = 3 + 1
=> x =4
\(\left(x-1\right)^2=9\)
\(\Leftrightarrow\left(x-1\right)^2=3^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3+1\\x=-1+3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}}\)
Vậy \(x=4\)hoặc \(x=2\)
\(\left(x-3\right)^3+15=7\)
\(\Leftrightarrow\left(x-3\right)^3=7-15\)
\(\Leftrightarrow\left(x-3\right)^3=-8\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow x-3=-2\)
\(\Leftrightarrow x=-2+3\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{2}+\dfrac{1}{4}\right)\times\left(9\times5-5\times9\right)=\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\times0=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Em nên viết rõ đề bài ra là 9\(^{x-1}\)hay là 9\(^x\)-1
Nếu là 9\(^{x-1}\)=1/9 thì 9\(^{x-1}\)=9\(^{-1}\)(vì 1/9=9\(^{-1}\))
Suy ra x-1=-1 nên x=0