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8 tháng 7 2021
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
=1,4 x\(\dfrac{15}{49}-\) \(\left(\dfrac{4}{5}+\dfrac{2}{3}\right)\) : 2\(\dfrac{1}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{22}{15}\) : \(\dfrac{11}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
( 2\(\dfrac{1}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(x\)) = \(\dfrac{3}{4}\)
\(\dfrac{11}{5}\) + \(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\)
\(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\) - \(\dfrac{11}{5}\)
\(\dfrac{3}{5}\)\(x\) = - \(\dfrac{29}{20}\)
\(x\) = -\(\dfrac{29}{12}\)
H
18 tháng 4 2023
`(x-5/12)^2 : 1 1/3= 0,25`
`=> (x-5/12)^2 : 4/3 = 0,25`
`=> (x-5/12)^2 =0,25 . 4/3`
`=> (x-5/12)^2 =1/3`
`=> (x-5/12)^2 =` \(\pm\left(\dfrac{1}{\sqrt{3}}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{12}=\dfrac{1}{\sqrt{3}}\\x-\dfrac{5}{12}=-\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{\sqrt{3}}+\dfrac{5}{12}\\x=-\dfrac{1}{\sqrt{3}}+\dfrac{5}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5+4\sqrt{3}}{12}\\x=\dfrac{5-4\sqrt{3}}{12}\end{matrix}\right.\)
`@ Yamada-Akiro`
26 tháng 8 2019
\(y+30\%y=-1,3\\ y\left(1-30\%\right)=-1,3\\ y\cdot0,7=-1,3\\ y=-1,3:0,7\\ y=\frac{-13}{7}\)
Vậy \(y=\frac{-13}{7}\)
\(y-0,25\%y=\frac{1}{2}\\ y\left(1-0,25\%\right)=0,5\\ y\cdot0,9975=0,5\\ y=0,5:0,9975\\ y=\frac{200}{399}\)
Vậy \(y=\frac{200}{399}\)
\(0,5x+\frac{2}{3}x=\frac{7}{12}\\ x\left(\frac{1}{2}+\frac{2}{3}\right)=\frac{7}{12}\\ x\cdot\frac{7}{6}=\frac{7}{12}\\ x=\frac{7}{12}:\frac{7}{6}\\ x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
\(\frac{\frac{3x}{7+1}}{-4}=\frac{-1}{8}\\ \Rightarrow8\cdot\frac{3x}{7+1}=4\\ \Rightarrow\frac{3x}{8}=\frac{1}{2}\\ \Rightarrow2\cdot3x=8\\ \Rightarrow6x=8\\ \Rightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)
chiều mai các bạn có được ko
Nobita Kun
HT
8 tháng 5 2017
Câu 1: 2 bài đầu hình như đề sai.
\(-\dfrac{5}{6}.\dfrac{4}{17}+-\dfrac{7}{12}.\dfrac{4}{7}\)
\(=\dfrac{4}{7}.\left(-\dfrac{5}{6}+-\dfrac{7}{12}\right)\)
\(=\dfrac{4}{7}.\dfrac{-17}{12}\)
\(=-\dfrac{1}{3}\)
Câu 2:
\(\dfrac{4}{5}.x=125\%-0,25\)
\(\dfrac{4}{5}.x=\dfrac{5}{4}-\dfrac{1}{4}\)
\(\dfrac{4}{5}.x=1\)
\(\Rightarrow\) \(x=1:\dfrac{4}{5}\)
\(\Rightarrow\) \(x=\dfrac{5}{4}\)
Vậy \(x=\dfrac{5}{4}\)
CC
5 tháng 7 2018
(0.25-0.3x)x1/3-1/4=-5/16
(0.25-0.3x)x1/3=-1/16 (làm tắt)
0.25-0.3x=-1/16:1/3
0.25-0.3x=-3/16
0.3x=0.4375 =>x=35/24
25 tháng 7 2018
(1/7+1/3) . x=28/3 . (1/4-1/7)
10/21 . x = 28/3 . 3/28
10/21 . x = 1
x = 1 : 10/21
x = 21/10
25 tháng 7 2018
\(\left(\frac{1}{7}+\frac{1}{3}\right).x=\frac{28}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\Leftrightarrow\frac{10}{21}.x=\frac{28}{3}.\frac{3}{28}\)
\(\Leftrightarrow\frac{10}{21}.x=1\)
\(\Leftrightarrow x=1:\frac{10}{21}\)
\(\Leftrightarrow x=\frac{21}{10}\)
Học tốt #
\(\left(\dfrac{x}{7}+0,25\right)-\dfrac{1}{3}=-\dfrac{1}{28}\)
\(\dfrac{x}{7}+\dfrac{1}{4}=-\dfrac{1}{28}+\dfrac{1}{3}\)
\(\dfrac{x}{7}+\dfrac{1}{4}=\dfrac{28}{84}+\dfrac{-3}{84}\)
\(\dfrac{x}{7}+\dfrac{1}{4}=\dfrac{25}{84}\)
\(\dfrac{x}{7}=\dfrac{25}{84}-\dfrac{1}{4}\)
\(\dfrac{x}{7}=\dfrac{25}{84}-\dfrac{21}{84}\)
\(\dfrac{x}{7}=\dfrac{4}{84}\)
\(\dfrac{x}{7}=\dfrac{1}{21}\)
=> x.21=1.7
x.21=7
x=7:21
\(x=\dfrac{1}{3}\)