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a)\(\left(\frac{1}{3}\right)^{-1}-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^4.2^3=3-1+\frac{1}{16}.8=3-1+\frac{1}{2}=\frac{5}{2}\\ \)
b)\(2^2.2^3.\left(\frac{2}{3}\right)^{-2}=2^5.\frac{9}{4}=72\)
c)\(\left(\frac{4}{3}\right)^{-2}.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^5:\left(\frac{3}{2}\right)^3=\frac{9}{128}\)
2)
\(3^{x+1}=9^x\Leftrightarrow3^x.3=9^x\Rightarrow3=9^x:3^x\Rightarrow3=3^x\Rightarrow x=1\)
\(\left(x-0,1\right)^2=6,25\Leftrightarrow\left(x-0,1\right)^2=2,5^2\Rightarrow\left(x-0,1\right)=2,5\Rightarrow x=2,5+0,1=2,6\)
\(3^{2x-1}=243\Leftrightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\Rightarrow x=1\)
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: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
\(1,a,\left(-\frac{3}{4}\right)^0=1\)
\(b,\left(-2\frac{1}{3}\right)^4=\left[-\left(\frac{2\cdot3+1}{3}\right)\right]^4=\left(\frac{-7}{3}\right)^4=\frac{2401}{256}\)
\(c,\left(2,5\right)^3=15,625\)
\(d,25^3:5^2=5^6:5^2=5^4=625\)
\(e,2^2\cdot4^3=2^2\cdot2^6=2^8=256\)
\(f,\left(\frac{1}{5}\right)^5\cdot5^3=\left(\frac{1}{5}\right)^5:\frac{1}{5^3}=\left(\frac{1}{5}\right)^5:\left(\frac{1}{5}\right)^3=\left(\frac{1}{5}\right)^2=\frac{1}{25}\)
\(g,\left(\frac{1}{5}\right)^3\cdot10^3=\left(\frac{1}{5}\cdot10\right)^3=2^3=8\)
\(h,\left(-\frac{2}{3}\right)^4:2^4=\left(-\frac{2}{3}:2\right)^4=\left(-\frac{1}{3}\right)^4=\frac{1}{81}\)
\(i,\left(\frac{2}{3}\right)^4:9^2=\left(\frac{2}{3}\right)^4:3^4=\left(\frac{2}{3}:3\right)^4=\left(\frac{2}{9}\right)^4=\frac{16}{6561}\)
\(k,\left(\frac{1}{2}\right)^3.\left(\frac{1}{4}^2\right)=\left(\frac{1}{2}\right)^3.\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^7=\frac{1}{128}\)
\(m,\left(\frac{120}{40}\right)^3=3^3=27\)
\(n,\frac{390^4}{130^4}=\left(\frac{390}{130}\right)^4=3^4=81\)
\(2,a,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}\cdot\frac{1}{2}\)
\(=2+\frac{1}{8}\)
\(=\frac{17}{8}\)
\(b,\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)
\(=-8+4+1+1\)
\(=-2\)
\(c,\left(3^2\right)^2-\left(\left(-5\right)^2\right)^2+\left(\left(-2\right)^3\right)^2\)
\(=3^4-\left(-5\right)^4+\left(-2\right)^6\)
\(=81-625+64\)
\(=-480\)
\(3,a,\left(x-1\right)^3=27\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
\(b,x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\Rightarrow x=-1\end{cases}}\)
\(c,\left(2x+1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1\Rightarrow x-1=\pm1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\text{ or }x=0\end{cases}}\)
\(4,M=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(M=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(M=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(M=100+99+98+97+...+2+1\)
\(M=\left(100+1\right)\cdot100:2\)
\(M=101\cdot50=5050\)
\(B=\left[\left(0,1\right)^2\right]^0+\left[\left(\frac{1}{7}\right)^{-1}\right]^2.\frac{1}{49}.\left[\left(2^2\right)^3.2^5\right]\)
\(=1+\left(\frac{1}{\frac{1}{7}}\right)^2.\frac{1}{49}.\left(2^6.2^5\right)\)
\(=1+7^2.\frac{1}{49}.2^{11}\)
\(\Rightarrow1+49.\frac{1}{49}.2^{11}\)
\(=1+2^{11}\)
Vậy \(B=1+2^{11}\)
\(\text{a,}\frac{2}{13}.\frac{-5}{3}+\frac{11}{13}.\frac{-5}{3}=-\frac{5}{3}\left(\frac{2}{13}+\frac{11}{13}\right)\)
\(=\frac{-5}{3}.\frac{13}{13}\)
\(=-\frac{5}{3}\)
\(\text{b,}\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^3.27+\left(\frac{-2017}{2018}\right)^0=\frac{1}{9}-\frac{1}{27}.27+1\)
\(=\frac{1}{9}-1+1\)
\(=\frac{1}{9}\)
\(\text{c,}1,2-\sqrt{\frac{1}{4}}:1\frac{1}{20}+\left|\frac{3}{4}-1,25\right|-\left(\frac{-3}{2}\right)^2=\frac{6}{5}-\frac{1}{2}:\frac{21}{20}+\left|\frac{3}{4}-\frac{5}{4}\right|-\frac{9}{4}\)
\(=\frac{6}{5}-\frac{10}{21}+\frac{1}{2}-\frac{9}{4}\)
\(=\frac{-431}{420}\)
a.
\(-2^3+2^2+\left(-1\right)^{2013}=-8+4-1=-5\)
b.
\(\left(3^3\right)^2-\left[\left(-2\right)^3\right]^2-\left(-5\right)^2=27^2-\left(-8\right)^2-25=729-64-25=640\)
c.
\(2^3+3\times\left(-\frac{1}{2016}\right)^0-\left(\frac{1}{2}\right)^2\times4-\left[\left(-2\right)^2\div\frac{1}{2}\right]=8+3\times0-\frac{1}{4}\times4-\left(4\times2\right)=8+3-1-8=2\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
a)
=(-1/1000)^2
=1/1000000
b)=(1)^3
=1