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Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
Bài 1:
a)
\((\frac{3}{5})^2-[\frac{1}{3}:3-\sqrt{16}.(\frac{1}{2})^2]-(10.12-2014)^0\)
\(=\frac{9}{25}-(\frac{1}{9}-1)-1\)
\(=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}\)
b)
\(|-\frac{100}{123}|:(\frac{3}{4}+\frac{7}{12})+\frac{23}{123}:(\frac{9}{5}-\frac{7}{15})\)
\(=\frac{100}{123}:\frac{4}{3}+\frac{23}{123}:\frac{4}{3}=(\frac{100}{123}+\frac{23}{123}):\frac{4}{3}=1:\frac{4}{3}=\frac{3}{4}\)
c)
\(\frac{(-5)^{32}.20^{43}}{(-8)^{29}.125^{25}}=\frac{5^{32}.(2^2.5)^{43}}{(-2)^{3.29}.(5^3)^{25}}=\frac{5^{32}.2^{86}.5^{43}}{-2^{87}.5^{75}}\)
\(=\frac{5^{32+43}.2^{86}}{-2^{87}.5^{75}}=\frac{5^{75}.2^{86}}{-2^{87}.5^{75}}=-\frac{1}{2}\)
Bài 2:
a)
\(\frac{2}{3}-(\frac{3}{4}-x)=\sqrt{\frac{1}{9}}=\frac{1}{3}\)
\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)
b)
\((\frac{1}{2}-x)^2=(-2)^2=2^2\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}-x=-2\\ \frac{1}{2}-x=2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=\frac{-3}{2}\end{matrix}\right.\)
c)
\(|3x+\frac{1}{2}|-\frac{2}{3}=1\)
\(|3x+\frac{1}{2}|=\frac{2}{3}+1=\frac{5}{3}\)
\(\Rightarrow \left[\begin{matrix} 3x+\frac{1}{2}=\frac{5}{3}\\ 3x+\frac{1}{2}=-\frac{5}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{7}{18}\\ x=\frac{-13}{18}\end{matrix}\right.\)
d)
\(3^{2x+1}=81=3^4\)
\(\Rightarrow 2x+1=4\Rightarrow x=\frac{3}{2}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)
d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)