1: Ta có: (x-47)-115=0
\(\Leftrightarrow x-47=115\)
\(\Leftrightarrow x=162\)(nhận)
Vậy: x=162
2: Ta có: \(2^x-15=17\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)(nhận)
Vậy: x=5
3: Ta có: \(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=800+200\)
\(\Leftrightarrow\left(7x-11\right)^3=1000\)
\(\Leftrightarrow\left(7x-11\right)^3=10^3\)
\(\Leftrightarrow7x-11=10\)
\(\Leftrightarrow7x=21\)
\(\Leftrightarrow x=3\)(nhận)
Vậy: x=3
4: Ta có: \(x^{10}=1^x\)
\(\Leftrightarrow x^{10}=1\)
\(\Leftrightarrow x^{10}=1^{10}\)
hay x=1(nhận)
Vậy: x=1
5: Ta có: \(x^{10}=x\)
\(\Leftrightarrow x^{10}-x=0\)
\(\Leftrightarrow x\left(x^9-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1\right\}\)
6) Ta có: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-15-1\right)\left(2x-15+1\right)=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\left(loại\right)\\x=8\left(nhận\right)\\x=7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{7;8\right\}\)
7: Ta có: \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
\(\Leftrightarrow2\cdot3^x=10\cdot3^{12}+3^{12}\cdot8\)
\(\Leftrightarrow2\cdot3^x=18\cdot3^{12}\)
\(\Leftrightarrow3^x=3^{12}\cdot9=3^{14}\)
hay x=14(nhận)
Vậy: x=14
1. (x - 47) - 115 = 0
x - 47 = 0 + 115
x - 47 = 115
x = 115 + 47
x = 162.
