1, \left(x + \dfrac{1}{2}\right)^2 = \dfrac{1}{9}
]
2, (3x - \dfrac{1}{4})^3 = -\dfrac{6}{4}
]
3, (x - \dfrac{1}{3})^2 + \left(x^2 - \dfrac{1}{9}\right)^4 = 0
]
4, 2x + 2 - 2x = 96 ]
5, 7x + 2 + 2 \cdot 7x - 1 = 345 ]
6, 7^x + 2 + 2 \cdot 7^{x - 1} = 345 ]
7, \left(\dfrac{1}{3}\right)^{2x - 1} = 243
]
8, (0.125)^{n - 1} = 64 ]
1: \(\left(x+\frac12\right)^2=\frac19\)
=>\(\left[\begin{array}{l}x+\frac12=\frac13\\ x+\frac12=-\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13-\frac12=-\frac16\\ x=-\frac13-\frac12=-\frac56\end{array}\right.\)
2: \(\left(3x-\frac14\right)^3=-\frac64\)
=>\(\left(3x-\frac14\right)^3=-\frac98\)
=>\(3x-\frac14=-\frac{\sqrt[3]{9}}{2}\)
=>\(3x=\frac14-\frac{\sqrt[3]{9}}{2}=\frac{1-2\cdot\sqrt[3]{9}}{4}\)
=>\(x=\frac{1-2\cdot\sqrt[3]{9}}{12}\)
4 Sửa đề: \(2^{x+2}-2^{x}=96\)
=>\(2^{x}\cdot4-2^{x}=96\)
=>\(2^{x}\cdot3=96\)
=>\(2^{x}=\frac{96}{3}=32=2^5\)
=>x=5
5: Sửa đề: \(7^{x+2}+2\cdot7^{x-1}=345\)
=>\(7^{x-1}\cdot7^3+2\cdot7^{x-1}=345\)
=>\(7^{x-1}\left(7^3+2\right)=345\)
=>\(7^{x-1}\left(343+2\right)=345\)
=>\(7^{x-1}=1\)
=>x-1=0
=>x=1
7: \(\left(\frac13\right)^{2x-1}=243\)
=>\(\left(\frac13\right)^{2x-1}=\left(\frac13\right)^{-5}\)
=>2x-1=-5
=>2x=-4
=>x=-2
8: \(0,125^{n-1}=64\)
=>\(\left(\frac18\right)^{n-1}=\left(\frac18\right)^{-2}\)
=>n-1=-2
=>n=-2+1=-1
