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13 tháng 8 2017

a) $\dfrac{120^3}{40^3}=(\dfrac{120}{40})^3=3^3=27$

b) $\dfrac{390^4}{130^4}=(\dfrac{390}{130})^4=3^4=81$

c) $\dfrac{3^2}{(0,375)^2}=(3:0,375)^2=(3:\dfrac{3}{8})^2=8^2=64$

6 tháng 8 2017

Giải:

a) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=30^3=2700\)

b) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=30^4=810000\)

c) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^2=8^2=64\)

Đáp số: a) 2700; b) 810000; c) 64.

Chúc bạn học tốt!!!

4 tháng 7 2017

a,\(\dfrac{120^3}{40^3}=3\)

b,\(\dfrac{390^3}{130^3}=3\)

c,\(\dfrac{3^2}{\left(0,375\right)^2}=8\)

8 tháng 7 2017

bn trình bày lời giải ra chứ

14 tháng 8 2017

image /assets/images/2017/08_10/1195-xHOKk8NXdv5EDVJo.jpeg

14 tháng 8 2017

a) \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)

b) \(\left(0,125\right)^3.512=\left(0,512\right)^3.8^3=\left(0,512.8\right)^3=1^3=1\)

c) \(\left(0,25\right)^4.1024=\left[\left(0,25\right)^2\right]^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)

d) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=3^3=64\)

e) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=3^4=81\)

g) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^3=8^3=512\)

17 tháng 7 2016

33 = 27

34= 81

82= 64

26 tháng 6 2019

1. sai dấu nhé 

2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)

b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)

c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)

Bài 1: Tìm x

a) Ta có: \(\dfrac{4}{3}:0.8=\dfrac{2}{3}:\left(0.1\cdot x\right)\)

\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}:\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow x\cdot\dfrac{1}{10}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\cdot\dfrac{3}{5}=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{2}{5}\cdot10=\dfrac{20}{5}=4\)

Vậy: x=4

b) Ta có: \(\left|x\right|=-1.2\)

mà \(\left|x\right|\ge0\forall x\)

nên \(x\in\varnothing\)

Vậy: \(x\in\varnothing\)

Bài 2: Tính

a) Ta có: \(\left(-2.5\right)\cdot\left(-4\right)\cdot\left(-7.9\right)\)

\(=\left(2.5\cdot4\right)\cdot\left(-7.9\right)\)

\(=-7.9\cdot10=-79\)

b) Ta có: \(\left(-0.375\right)\cdot\dfrac{13}{3}\cdot\left(-2\right)^3\)

\(=\dfrac{3}{8}\cdot8\cdot\dfrac{13}{3}\)

\(=3\cdot\dfrac{13}{3}=13\)

14 tháng 5 2021

\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)

\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)

\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)

14 tháng 5 2021

a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)

\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)

\(=\dfrac{2}{5}+-2\)

\(=\dfrac{2}{5}+\dfrac{-10}{5}\)

\(=\dfrac{-8}{5}\)

6 tháng 8 2021

a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)

c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)

6 tháng 8 2021

A = 2/3 + -1/3

    = 1/3

B = 25/11 . 13/12 . (-2,2)

    = 325/132 . (-2,2)

    = -65/12

C = 11/20 . -2/5

    = -11/50

Chúc bạn học tốt!! ^^

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