a) \(-0.8\sqrt{\left(-0.125\right)^2}=-0.8\left|-0.125\right|=-0.8\times0.125=0,1\)

b) \(\sqrt{\left(-2\right)^6}=\sqrt{2^6}=\sqrt{\left(2^3\right)^2}=\left|8\right|=8\)

a) \(-0,8\sqrt{\left(-0,125\right)^2}=-0,8.\left|-0,125\right|=-0.8.0,125=-\dfrac{1}{10}\)

b) \(\sqrt{\left(-2\right)^6}=\sqrt{\left(\left(-2\right)^3\right)^2}=\left|\left(-2\right)^3\right|=8\)

c) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)

d) \(\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

\(A=\left[0,8\cdot7+(0,8)^2\right]\cdot\left[1,25\cdot7-\frac{4}{5}\cdot1,25\right]-47,86\)

\(=0,8\cdot(7+0,8)\cdot1,25\cdot(7-0,8)-47,86\)

\(=0,8\cdot7,8\cdot1,25\cdot6,2-47,86\)

\(=48,36-47,86=0,5\)

\(B=\frac{(1,09-0,29)\cdot\frac{5}{4}}{(18,9-16,65)\cdot\frac{8}{9}}=\frac{0,8\cdot1,25}{2,25\cdot\frac{8}{9}}=\frac{1}{2}\)

\(A:B=0,5:\frac{1}{2}=\frac{1}{2}:\frac{1}{2}=\frac{1}{2}\cdot2=1\)

A gấp 1 lần B

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\(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{20^2}=\dfrac{17}{4800}\)

b) \(\left(-2\dfrac{3}{4}+\dfrac{1}{2}\right)^2\)

\(=\left(\dfrac{-11}{4}+\dfrac{1}{2}\right)^2\)

\(=\left(\dfrac{-11}{4}+\dfrac{2}{4}\right)^2\)

\(=\left(\dfrac{-9}{4}\right)^2\)

\(=\dfrac{81}{16}\)

Giúp mình với nha T.T

a) x = \(\dfrac{-64}{3}\)

b) x = -3,5

c) x = 80

d) x = -1.162

e) x = 0,9436

g) x \(\in\varnothing\)

a) 16/3 : x = -1/4

=> x = 16/3 : (-1/4)

=> x = 16/3 . (-4)

=> x = -64/3

Vậy x= -64/3

b)2x - 1³ = -8

=> 2x = (-8) + 1

=> 2x = -7

=> x = -7/2

d) 0,944 - 2x = 3,268

=> 2x = 0,944 - 3,268

=> 2x = -2,324

=> x = (-2,324) : 2

=> x = -1,162

g) \(\sqrt{5^2-3^2}=-\sqrt{81-x}\)

=> \(\sqrt{25-9}\)= \(-\sqrt{81-x}\)

=> \(\sqrt{16}\)=\(-\sqrt{81-x}\)

=> 4=\(-\sqrt{81-x}\)

tới đây mik bí r hk bt lm nữa