\(\left|x\right|-3=4\)
\(\Rightarrow\left|x\right|=4+3\)
\(\Rightarrow\left|x\right|=7\)
\(\Rightarrow x\in\left\{-7;7\right\}\)
\(\frac{10^3+2.5^3+5^3}{55}=\frac{2^3.5^3+5^3\left(1+2\right)}{5.11}=\frac{8.5^3+5^3.3}{5.11}=\frac{5^3\left(8+3\right)}{5.11}\)
\(=\frac{5^3.11}{5.11}=\frac{5^3}{5}=5^2=25\)
Bài 1 :
\(\left|x\right|-3=4\)
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
Vậy...
Bài 2 :
\(\frac{10^3+2\cdot5^3+5^3}{55}\)
\(=\frac{2^3\cdot5^3+2\cdot5^3+5^3}{5\cdot11}\)
\(=\frac{5^3\cdot\left(2^3+2+1\right)}{5\cdot11}\)
\(=\frac{5^3\cdot11}{5\cdot11}\)
\(=5^2\)