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a. \(25^3:5^2\)
\(=\left(5^2\right)^3:5^2\)
\(=5^6:5^2=5^4\)
b. \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21-\left(2+6\right)}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)
\(a,25^3:5^2\)
=\(\left(5^2\right)^3:5^2\)
=\(5^6:5^2\)
=\(5^4\)
\(b,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
=\(\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^9\)
\(c,3-\left(\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
=\(3-1+\frac{1}{4}:2\)
\(=2+\frac{1}{4}\cdot\frac{1}{2}\)
\(=2+\frac{1}{8}\)
\(=\frac{17}{8}\)
\(d,\left(-\frac{7}{4}:\frac{5}{8}\right)\cdot\frac{11}{16}\)
\(=\left(-\frac{7}{4}\cdot\frac{8}{5}\right)\cdot\frac{11}{16}\)
\(=-\frac{14}{5}\cdot\frac{11}{16}\)
\(=-\frac{77}{40}\)
\(e,\frac{2}{3}+\frac{1}{3}\cdot\frac{-6}{10}\)
\(=\frac{2}{3}-\frac{1}{5}\)
\(=\frac{7}{15}\)
Bài 1:
\(a,A=\frac{-25}{28}.0,21=\frac{-25}{28}.\frac{21}{100}=\frac{-25.21}{28.100}=\frac{-1.25.3.7}{4.7.25.4}=\frac{-1.3}{4.4}=\frac{-3}{16}\)
\(b,B=\left(\frac{13}{24}-\frac{29}{30}\right):\left(-10,2\right)=\left(\frac{65}{120}-\frac{116}{120}\right):\frac{-51}{5}=\frac{-51}{120}.\frac{5}{-51}=\frac{-51.5}{120.\left(-51\right)}=\frac{-51.5}{5.24.\left(-51\right)}=\frac{1}{24}\)
\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{99}{100}-\frac{99}{100}\)
\(A=\frac{99-99}{100}=0\)
Bài 2
\(\left(3x+5\right).\left(2x-4\right)=0\)
\(TH1:3x+5=0\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(TH2:2x-4=0\)
\(2x=4\)
\(x=2\)
\(\left(x^2-1\right).\left(x+3\right)=0\)
\(\Rightarrow x^2-1=0\)
\(x^2=1\)
\(\Rightarrow x=1\)
\(x+3=0\)
\(x=-3\)
\(5x^2-\frac{1}{2}x=0\)
\(\Rightarrow5x^2-\frac{x}{2}=0\)
\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)
\(10x^2-x=x.\left(10x-1\right)\)
\(\frac{x.\left(10x-1\right)}{2}=0\)
\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)
\(10x-1=0\)
\(x=\frac{1}{10}=0.100\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)
\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)
\(\frac{x}{4}=\frac{5}{4}\)
\(\Rightarrow x=5\)
\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)
\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)
\(x=\frac{7}{8}:\frac{5}{8}\)
\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)
#)Well ! Bài này cg dạng tầm cỡ vừa :v
Làm cho ai v ? mau vô nhận bài đê !
ờ tụi làm cho bạn trên lớp í mà
Mà tui ngu quá nhóm lun từ số 1 nhanh hơn ko
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: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
\(\text{a,}\frac{2}{13}.\frac{-5}{3}+\frac{11}{13}.\frac{-5}{3}=-\frac{5}{3}\left(\frac{2}{13}+\frac{11}{13}\right)\)
\(=\frac{-5}{3}.\frac{13}{13}\)
\(=-\frac{5}{3}\)
\(\text{b,}\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^3.27+\left(\frac{-2017}{2018}\right)^0=\frac{1}{9}-\frac{1}{27}.27+1\)
\(=\frac{1}{9}-1+1\)
\(=\frac{1}{9}\)
\(\text{c,}1,2-\sqrt{\frac{1}{4}}:1\frac{1}{20}+\left|\frac{3}{4}-1,25\right|-\left(\frac{-3}{2}\right)^2=\frac{6}{5}-\frac{1}{2}:\frac{21}{20}+\left|\frac{3}{4}-\frac{5}{4}\right|-\frac{9}{4}\)
\(=\frac{6}{5}-\frac{10}{21}+\frac{1}{2}-\frac{9}{4}\)
\(=\frac{-431}{420}\)
\(\left|\frac{-1}{10}\right|-\left(\frac{1}{2}\right)^2\div\frac{5}{9}\)
=\(\frac{1}{10}-\frac{1}{4}\times\frac{9}{5}\)
=\(\frac{1}{10}-\frac{9}{20}\)
=\(\frac{1\times2-9}{20}\)=\(\frac{-7}{20}\)