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1: \(=\left(\dfrac{1}{3}\right)^{25}\cdot90^{25}\cdot\dfrac{1}{3^{25}}-\dfrac{2}{12}\)

\(=\dfrac{30^{25}}{3^{25}}-\dfrac{1}{6}=10^{25}-\dfrac{1}{6}\)

2: \(=10\cdot1\cdot\dfrac{4}{3}+7-\dfrac{1}{6}\cdot2=20\)

25 tháng 10 2022

a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)

b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)

=10+3/64

=643/64

c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)

a: \(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\cdot12=20\)

b: \(=\dfrac{3}{5}:\left(\dfrac{-2-5}{30}\right)+\dfrac{3}{5}:\left(\dfrac{-1}{3}-\dfrac{16}{15}\right)\)

\(=\dfrac{3}{5}:\dfrac{-7}{30}+\dfrac{3}{5}:\dfrac{-21}{15}\)

\(=\dfrac{3}{5}\left(\dfrac{-30}{7}-\dfrac{15}{21}\right)=\dfrac{3}{5}\cdot\left(\dfrac{-30}{7}-\dfrac{5}{7}\right)=\dfrac{3}{5}\cdot\left(-5\right)=-3\)

c: \(=5.7\left(-6.5-3.5\right)\)

\(=5.7\cdot\left(-10\right)=-57\)

d: \(=10\cdot0.1\cdot\dfrac{4}{3}+3\cdot7-\dfrac{1}{6}\cdot2\)

\(=\dfrac{4}{3}+21-\dfrac{1}{3}=22\)

25 tháng 10 2022

a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)

b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)

\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)

c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)

 

21 tháng 9 2017

a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)

=1+3+5+7+9

=25

b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)

=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)

=\(\dfrac{15}{12}\)

c) =0,2+0.3+0,4

= 0.9

d) =9-8+7

=8

j) =1,2-1,3+1.4

= (-0,1)+1,4

=1,4

g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)

= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)

= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)

=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)

= \(\dfrac{71}{20}\)

Nhớ tick cho mk nha~

11 tháng 2 2018

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)

\(B=\frac{300}{343}:\frac{1347}{343}\)

\(B=\frac{100}{449}\)

11 tháng 2 2018

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)

\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)

\(A=\frac{-1}{2}+\frac{1710}{9}\)

\(A=\frac{-1}{2}+190\)

\(A=\frac{-1}{2}+\frac{380}{2}\)

\(A=\frac{379}{2}\)