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GH
1
VN
20 tháng 12 2022
a)
\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{18}{24}-\dfrac{20}{24}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-12\times\dfrac{144}{1}\)
\(=-1728\)
b)
\(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(4\times\dfrac{3}{4}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(3-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\times\dfrac{6}{5}-17\)
\(=3-17\)
\(=-14\)
S
20 tháng 12 2022
a)\(=-12:\left(-\dfrac{1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)\(=-12.144=-1728\)
b)\(=\left(8:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\left(6-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\dfrac{11}{2}.\dfrac{6}{5}-17=\dfrac{33}{5}-17=\dfrac{33}{5}-\dfrac{85}{5}=-\dfrac{2}{5}\)
TH
4
TH
2 tháng 10 2021
GIÚP MÌNH VỚI MN ƠIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
2 tháng 10 2021
\(A=\dfrac{\left(20.5\right)^5.5^5}{100^5}=\dfrac{100^5.3125}{100^5}=3125\)
\(B=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^5.0,3}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^5.0,3}=\dfrac{3^5}{0,3}=810\)
12 tháng 11 2021
c: \(=\dfrac{3}{2}\cdot1-1-20=\dfrac{3}{2}-21=\dfrac{-39}{2}\)
28 tháng 7 2023
a: \(=6+\dfrac{4}{5}-1-\dfrac{2}{3}-3-\dfrac{4}{5}\)
\(=2-\dfrac{2}{3}=\dfrac{4}{3}\)
b: \(=7+\dfrac{5}{9}-2-\dfrac{3}{4}-3-\dfrac{5}{9}=2-\dfrac{3}{4}=\dfrac{5}{4}\)
c: =6+7/7-1-3/4-2-5/7
=3+2/7-3/4
=84/28+8/28-21/28
=84/28-13/28
=71/28
6 tháng 2 2022
\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{3^4\left(3-1\right)}{3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)
\(=\dfrac{1}{3\cdot2}-\dfrac{1}{5}\cdot\dfrac{-6}{9}=\dfrac{1}{6}+\dfrac{6}{45}=\dfrac{45+36}{270}=\dfrac{81}{270}=\dfrac{3}{10}\)
4 tháng 12 2021
\(a,=5\cdot0,6-10\cdot0,2=3-2=1\\ b,=\dfrac{1}{9}:\left(\dfrac{1}{30}\right)^2=\dfrac{1}{9}:\dfrac{1}{900}=\dfrac{1}{9}\cdot900=100\)
17 tháng 12 2023
a: \(\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
\(=\left(\dfrac{3^6}{9}-81\right)\left(\dfrac{3}{4}-81\right)\cdot\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
\(=\left(81-81\right)\left(\dfrac{3}{4}-81\right)\cdot\left(\dfrac{3^2}{5}-81\right)\cdot...\cdot\left(\dfrac{3^{2000}}{2003}-81\right)\)
=0
b: \(\dfrac{69}{157}-\left(2+\left(3+4+5^{-1}\right)^{-1}\right)^{-1}\)
\(=\dfrac{69}{157}-\left(2+\left(3+4+\dfrac{1}{5}\right)^{-1}\right)^{-1}\)
\(=\dfrac{69}{157}-\left(2+1:\dfrac{36}{5}\right)^{-1}\)
\(=\dfrac{69}{157}-\left(2+\dfrac{5}{36}\right)^{-1}\)
\(=\dfrac{69}{157}-\left(\dfrac{77}{36}\right)^{-1}\)
\(=\dfrac{69}{157}-\dfrac{36}{77}=\dfrac{-339}{12089}\)
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
A=4+16+36+64+100+144+196+256+324+400
A=20+100+100+340+580+400
A=320+920+400
A=1240+400
A=1640
\(A=2^2+4^2+6^2+...+20^2.\)
\(=2^2\left(1+2^2+3^2+...+10^2\right)\)
\(=2^2\left[1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+10\left(11-1\right)\right]\)
\(=2^2\left[\left(1.2+2.3+3.4+...+10.11\right)-\left(1+2+3+...+10\right)\right]\)
\(=2^2\left[\frac{1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+10.11.\left(12-9\right)}{3}-\frac{10\left(10+1\right)}{2}\right]\)
\(=2^2\left[\frac{1.2.3-0.1.2+2.3.4-1.2.3+...+10.11.12-9.10.11}{3}-55\right]\)
\(=2^2\left(\frac{10.11.12}{3}-55\right)=2^2.275=1100\)