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a) \(x+\frac{1}{2}=2^5:2^3\)
\(x+\frac{1}{2}=4\)
\(x=4-\frac{1}{2}\)
\(x=\frac{7}{2}\)
b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
\(\frac{5}{3}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{5}{3}\)
\(x=\frac{1}{35}\)
c)\(\left|x+5\right|-6=9\)
\(\left|x+5\right|=9+6\)
\(\left|x+5\right|=15\)
\(\Rightarrow\orbr{\begin{cases}x+5=15\\x+5=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=15-5\\x=\left(-15\right)-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-20\end{cases}}}\)
Vậy x = 10 ; x= -20
d)\(-\frac{12}{13}x-5=6\frac{1}{3}\)
\(-\frac{12}{13}x-5=\frac{19}{3}\)
\(-\frac{12}{13}x=\frac{19}{3}+5\)
\(-\frac{12}{13}x=\frac{34}{3}\)
\(x=\frac{34}{3}:\left(-\frac{12}{13}\right)\)
\(x=-\frac{221}{18}\)
c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)
1.
\(\left(x+2\right)^3=\frac{1}{8}\)
\(\Rightarrow\left(x+2\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow x+2=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}-2\)
\(\Rightarrow x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}.\)
2.
b) Ta có:
\(5^5-5^4+5^3\)
\(=5^3.\left(5^2-5+1\right)\)
\(=5^3.\left(25-5+1\right)\)
\(=5^3.21\)
Vì \(21⋮7\) nên \(5^3.21⋮7.\)
\(\Rightarrow5^5-5^4+5^3⋮7\left(đpcm\right).\)
c) Ta có:
\(2^{19}+2^{21}+2^{22}\)
\(=2^{19}.\left(1+2^2+2^3\right)\)
\(=2^{19}.\left(1+4+8\right)\)
\(=2^{19}.13\)
Vì \(13⋮13\) nên \(2^{19}.13⋮13.\)
\(\Rightarrow2^{19}+2^{21}+2^{22}⋮13\left(đpcm\right).\)
Chúc bạn học tốt!