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\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
Bài 1:
a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{-2}{2}=-1\)
b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)
\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)
\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)
\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)
c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)
\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)
d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)
\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)
\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)
e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)
\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)
\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)
\(=\frac{11}{12}\)
f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
a) \(\left|\frac{1}{3}x-8\right|+3=15\)
\(\Leftrightarrow\left|\frac{1}{3}x-8\right|=12\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x-8=-12\\\frac{1}{3}x-8=12\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x=-4\\\frac{1}{3}x=20\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-12\\x=60\end{cases}}\)
Vậy \(x\in\left\{-12;60\right\}\)
b) \(15-\left|2+3x\right|=8\)
\(\Leftrightarrow\left|2+3x\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=-7\\2+3x=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=-9\\3x=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{5}{3}\end{cases}}\)
Vậy \(x\in\left\{-3;\frac{5}{3}\right\}\)
d) \(-1\frac{1}{6}-\left|5-3x\right|=\frac{2}{3}\)
\(\Leftrightarrow\frac{-7}{6}-\left|5-3x\right|=\frac{2}{3}\)
\(\Leftrightarrow\left|5-3x\right|=\frac{-7}{6}-\frac{2}{3}\)
\(\Leftrightarrow\left|5-3x\right|=\frac{-11}{6}\)
Vì \(\left|5-3x\right|\ge0\forall x\)
mà \(\frac{-11}{6}< 0\)\(\Rightarrow\)Vô lý
Vậy \(x\in\varnothing\)
e) \(\left(\frac{3}{7}\right)^{20}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{20}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{20}:\left(\frac{3}{7}\right)^{2.6}\)
\(=\left(\frac{3}{7}\right)^{20}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^8\)
g) \(4.2^5:\left(2^3.1^{16}\right)=2^2.2^5:2^3=2^4=16\)