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b) = \(\frac{3}{4}\div\)\(\left(-\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\)
= \(\frac{3}{4}\div\frac{5}{6}\)
= \(\frac{9}{10}\)
c) \(\frac{16.2^3}{4}\)
\(=4.8=32\)
\(a)\left|-\frac{1}{2}\right|+3^0+\frac{1}{4}+4+2021^0.\)
\(=\frac{1}{2}+1+\frac{1}{4}+4+1\)
\(=\left(\frac{1}{2}+\frac{1}{4}\right)+\left(1+4+1\right)\)
\(=\frac{3}{4}+6=\frac{27}{4}\)
\(b)\frac{3}{4}\div\left(-\frac{1}{3}\right)+\frac{3}{4}\div\frac{2}{3}+\frac{3}{4}\div\frac{1}{2}\)
\(=\frac{3}{4}\div\left(-\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\)
\(=\frac{3}{4}\div\frac{5}{6}=\frac{9}{10}\)
1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
\(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
\(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
\(2x=\frac{53}{30}\)
\(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
\(2x=\frac{37}{30}\)
\(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
\(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
\(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
\(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
\(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
\(-\frac{5}{7}x=-\frac{11}{45}\)
\(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}
A= E387E4837
B = 883433
C = UỲUWFHQWURY48E3947
a) \(\frac{\left(0,8\right)^2}{\left(0,4\right)^2}=4\)
b) 754 : 254 + 20120
= (75:25)4 + 1
= 34 + 1 = 81 + 1 = 82
a) \(\frac{\left(0,8\right)^2}{\left(0,4\right)^2}=\left(\frac{0,8}{0,4}\right)^2=2^2=4\)
b) \(75^4:25^4+2012^0\)
\(=\left(75:25\right)^4+2012^0\)
\(=3^4+2012^0\)
\(=81+1\)
\(=82\)
a) Ta có: \(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+\frac{1}{2}\cdot8-5+64\)
\(=-8+4-5+64=55\)
b) Ta có: \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}\right)\cdot\frac{3}{2}+\left(\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=0\cdot\frac{3}{2}=0\)
c) Ta có: \(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\frac{2^{10}\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}=\frac{2\left(2^9\cdot9^4-6^9\right)}{6^8\left(2^2+20\right)}=\frac{-1}{3}\)
a) ( -2 )3 + \(\frac{1}{2}:\frac{1}{8}\) - √25 + \(|-64|\)
= \(\frac{-8}{1}\) + \(\frac{1}{2}.\frac{8}{1}\) - \(\frac{5}{1}\) + \(\frac{64}{1}\)
= \(\frac{-16}{2}+\frac{1}{2}.\frac{8}{1}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16}{2}+\frac{8}{2}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16+8-10+128}{2}\) = \(\frac{110}{2}\) = 55
\(\left(x+\frac{1}{2}\right)\left(x-\frac{3}{4}\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)
a)\(\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)
\(=\left(-8\right)+4+1+1\)
\(=-2\)
b)\(2^4+8\left(\frac{-2^2}{\frac{1}{2}}\right)^0-2^{-2}.4+\left(-2\right)^2\)
\(=16+8.1-\frac{1}{2^2}.4+4\)
\(=16+8-\frac{1}{4}.4+4\)
\(=16+8-4+4\)2
\(=24\)
Có thể bạn không hiểu một số chỗ nên mình giải thích luôn:
\(-1^a\)có thể là 1 nếu số a là chẵn hoặc -1 nếu số a là lẻ. Vd:\(\left(-1\right)^3=-1,\left(-1\right)^2=1\)
Một số với mũ 0 = 1 (a0=1), với a là số. Vd: \(2^0=1\)
Một số với mũ âm = \(\frac{1}{a^n}\)(với a là số, n là số mũ âm bỏ dấu trừ) Vd:\(2^{-2}=\frac{1}{2^2}\)
kéo đồng đội đến cuau
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