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19 tháng 1 2017

\(\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{2}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)

\(=\frac{100}{123}:\frac{17}{4}+\frac{23}{123}:\frac{4}{3}\)

\(=\frac{100}{123}\times\frac{4}{17}+\frac{23}{123}\times\frac{3}{4}\)

\(=\frac{400}{2091}+\frac{23}{164}=\frac{2773}{8364}\)

Thanks you

1 tháng 12 2019

\(a,\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2=\left(\frac{2}{2.3}\right)^6.\left(\frac{5}{2}\right)^4\)

\(=\frac{1}{3^6}.\frac{5^4}{2^4}=\frac{5^4}{3^6.2^4}\)

\(b,\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)

\(=\frac{100}{123}:\left(\frac{9+7}{12}\right)+\frac{23}{123}:\left(\frac{27-7}{15}\right)\)

\(=\frac{100}{123}:\frac{16}{12}+\frac{23}{123}:\frac{20}{15}\)

\(=\frac{100.12}{123.16}+\frac{23.15}{123.20}\)

\(=\frac{5.5.4.3.4}{41.3.4.4}+\frac{23.3.5}{41.3.4.5}\)

\(=\frac{25}{41}+\frac{23}{164}=\frac{25.4+23}{164}\)

\(=\frac{123}{164}=\frac{3}{4}\)

Bài 1

\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)

\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)

\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)

\(=\frac{9}{25}+\frac{8}{9}-1\)

\(=\frac{56}{225}\)

\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)

\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)

\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)

\(=1:\frac{4}{3}=\frac{3}{4}\)

Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v 

\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)

\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)

\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)

\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)

\(=-\frac{1}{2}\)

AH
Akai Haruma
Giáo viên
1 tháng 1 2020

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)

\(\frac{(-5)^{32}.20^{43}}{(-8)^{29}.125^{25}}=\frac{5^{32}.(2^2.5)^{43}}{(-2)^{3.29}.(5^3)^{25}}=\frac{5^{32}.2^{86}.5^{43}}{-2^{87}.5^{75}}\)

\(=\frac{5^{32+43}.2^{86}}{-2^{87}.5^{75}}=\frac{5^{75}.2^{86}}{-2^{87}.5^{75}}=-\frac{1}{2}\)

AH
Akai Haruma
Giáo viên
1 tháng 1 2020

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: \(A=\dfrac{63-48-70}{84}:\dfrac{-3\cdot84+4\cdot60+5\cdot70}{840}\cdot\dfrac{1-14}{12}\)

\(=\dfrac{-55}{84}\cdot\dfrac{840}{338}\cdot\dfrac{-13}{12}=\dfrac{55}{1}\cdot\dfrac{10}{338}\cdot\dfrac{13}{12}=\dfrac{275}{156}\)

b: \(=-234\cdot123+123\cdot4356-123\cdot2312+234\cdot123-234\cdot2312+2312\cdot234+2312\cdot123\)

\(=123\cdot4356-123\cdot2312+123\cdot2312=123\cdot4356=535788\)

4 tháng 10 2021

yutyugubhujyikiu

23 tháng 9 2016

bai de the ma cung hoi