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12 tháng 7 2023

\(\dfrac{1}{2}-\left(40\%-0,75\right)\)

\(=\dfrac{1}{2}-\left(\dfrac{40}{100}-\dfrac{3}{4}\right)\)

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

\(=\dfrac{1}{2}-\left(\dfrac{8}{20}-\dfrac{15}{20}\right)\)

\(=\dfrac{1}{2}-\left(-\dfrac{7}{20}\right)\)

\(=\dfrac{1}{2}+\dfrac{7}{20}\)

\(=\dfrac{10}{20}+\dfrac{7}{20}\)

\(=\dfrac{17}{20}\)

12 tháng 7 2023

cảm ơn bn yeu

 

AH
Akai Haruma
Giáo viên
27 tháng 11 2023

Lời giải:

$\frac{1}{4}-3x+\frac{3}{2}=-0,75$

$3x=\frac{1}{4}+\frac{3}{2}-(-0,75)=2,5$

$\Rightarrow x=2,5:3=\frac{5}{6}$

(-5/24+0,75+7/12).(-2 1/4)

=(-5/24+3/4+7/12).(-9/4)

=9/8.(-9/4)

=-81/32

1.110.05

2.2560m2

10 tháng 8 2023

a) \(0,6+\dfrac{2}{3}=\dfrac{6}{10}+\dfrac{2}{3}=\dfrac{3}{5}+\dfrac{2}{3}=\dfrac{9}{15}+\dfrac{10}{15}=\dfrac{19}{15}\)

b) \(-\dfrac{5}{12}+0,75=-\dfrac{5}{12}+\dfrac{75}{100}=-\dfrac{5}{12}+\dfrac{3}{4}=-\dfrac{5}{12}+\dfrac{9}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)

c) \(\dfrac{1}{3}-\left(-0,4\right)=\dfrac{1}{3}+\dfrac{4}{10}=\dfrac{1}{3}+\dfrac{2}{5}=\dfrac{5}{15}+\dfrac{6}{15}=\dfrac{11}{15}\)

d) \(1\dfrac{3}{5}+\dfrac{5}{6}=\dfrac{8}{5}+\dfrac{5}{6}=\dfrac{48}{40}+\dfrac{25}{30}=\dfrac{73}{30}\)

18 tháng 6

a; (15 - \(\dfrac{121}{18}\)) : \(\dfrac{297}{27}\) - \(\dfrac{17}{8}\) : \(\dfrac{51}{40}\)

     (\(\dfrac{270}{18}\) - \(\dfrac{121}{18}\)) : \(\dfrac{297}{27}\) - \(\dfrac{17}{8}\) x \(\dfrac{40}{51}\)

 =   \(\dfrac{149}{18}\) : \(\dfrac{297}{27}\) - \(\dfrac{5}{3}\)

\(\dfrac{149}{18}\) x \(\dfrac{27}{297}\) - \(\dfrac{5}{3}\)

 = \(\dfrac{149}{198}\) - \(\dfrac{5}{3}\)

\(\dfrac{149}{198}\) - \(\dfrac{330}{198}\)

\(\dfrac{-181}{198}\)

18 tháng 6

b; (- 3,2) x (- \(\dfrac{15}{64}\)) + (0,8 - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)

= (\(\dfrac{-16}{5}\)) x ( \(\dfrac{-15}{64}\)) + (\(\dfrac{4}{5}\) - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)

\(\dfrac{3}{4}\) + (\(\dfrac{4}{5}\) - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)

\(\dfrac{3}{4}\) + (\(\dfrac{12}{15}\) - \(\dfrac{34}{15}\)) : \(\dfrac{11}{3}\)

\(\dfrac{3}{4}\) + \(\dfrac{-22}{15}\) : \(\dfrac{11}{3}\)

\(\dfrac{3}{4}\) - \(\dfrac{22}{15}\) x \(\dfrac{3}{11}\)

\(\dfrac{3}{4}\) - \(\dfrac{2}{5}\)

\(\dfrac{15}{20}\) - \(\dfrac{8}{20}\)

\(\dfrac{7}{20}\)

4 tháng 5 2019

A, \(2\frac{2}{5}\left(\frac{1}{2}x-0,75\right)=\frac{3}{10}\)

\(=>\frac{2.5+2}{5}\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)

\(=>\frac{1}{2}x-\frac{3}{4}=\frac{3}{10}:\frac{12}{5}=\frac{1}{8}\)

\(=>x=\left(\frac{1}{8}+\frac{3}{4}\right):\frac{1}{2}\)

\(=>x=\frac{7}{4}\)

B, \(\frac{3}{5}-|x-\frac{1}{2}|=25\%\)

\(=>|x-\frac{1}{2}|=\frac{3}{5}-\frac{1}{4}\)

\(=>|x-\frac{1}{2}|=\frac{7}{20}\)

\(=>x-\frac{1}{2}=\frac{7}{20};-\frac{7}{20}\)

TH1: \(x-\frac{1}{2}=\frac{7}{20}=>x=\frac{17}{20}\)

TH2: \(x-\frac{1}{2}=-\frac{7}{20}=>x=\frac{3}{20}\)

a. 8,5 x 202,1 + 1,5 x 202,1 + 202,1 x 0,75 + 0,25 x 202,1 - 202,1 

= 202,1 x ( 8,5 + 1,5 + 0,75 + 0,25 - 1) 

= 202,1 x 10 

= 2021 

b. ( 0,8 + 0,5 + 0,75 + 0,5 + 0,2 + 0,25) : 0,5 

= 3 : 0,5 

= 6 

 

c. 0,4 x 0,75 + 0,4 + 0,4 x 0,5 x 0,4 x 0,25 + 0,5 x 0,4 - 0,4 

= 0,4 x ( 0,75 + 1 + 0,05 + 0,5 - 1) 

= 0,4 x 1,3 

= 0,52

=(1/2-3/4)*(1/5-2/5):5/9-3/2

=-1/4*(-1/5)*9/5-3/2

=1/20*9/5-3/2

=9/100-3/2=9/100-150/100=-141/100