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9 tháng 1 2022
a: \(\Leftrightarrow-2x=-\dfrac{1}{3}-\dfrac{1}{8}-\dfrac{5}{7}=-\dfrac{197}{168}\)
hay x=197/336
c: \(\Leftrightarrow5x=9+\dfrac{6}{18}-\dfrac{2}{7}=\dfrac{190}{21}\)
hay x=38/21
NT
1
6 tháng 7 2018
\(2.\left(\dfrac{3}{4}-5x\right)=\dfrac{4}{5}-3x\)
= \(2.\dfrac{3}{4}-2.5x=\dfrac{4}{5}-3x\)
= \(\dfrac{3}{2}-10x=\dfrac{4}{5}-3x\)
= \(\dfrac{3}{2}-\dfrac{4}{5}=10x-3x\)
\(\dfrac{7}{10}\) = 7x
\(\dfrac{7}{10}:7=x\)
\(\dfrac{1}{10}=x\)
24 tháng 7 2021
a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)
\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)
\(\dfrac{19}{20}x=1\)
\(x=\dfrac{20}{19}\)
Vậy \(x=\dfrac{20}{19}\)
24 tháng 7 2021
b) \(\left(x^2-9\right)\left(3-5x\right)=0\)
TH1:
\(x^2-9=0\)
\(x^2=9\)
\(x^2=3^2=\left(-3\right)^2\)
=>\(x\in\left\{3;-3\right\}\)
TH2:
\(3-5x=0\)
\(5x=3\)
\(x=\dfrac{3}{5}\)
Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)
CT
6 tháng 5 2018
\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)
=> \(x=56\)
2) => 18x = 18
=> x = 1
3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)
=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)
=> \(x=\dfrac{-1}{2}\)
4) 45%.x =\(\dfrac{3}{5}\)
=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)
=> \(x=\dfrac{4}{3}\)
17 tháng 7 2021
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
17 tháng 7 2021
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
QN
2
VD
20 tháng 3 2022
\(a,\dfrac{8}{14}=\dfrac{x+1}{7}\\ \Leftrightarrow\dfrac{x+1}{7}=\dfrac{4}{7}\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\\ b,\dfrac{-4}{3}=\dfrac{5x+1}{-27}\\ \Leftrightarrow\dfrac{36}{-27}=\dfrac{5x+1}{-27}\\ \Leftrightarrow5x+1=36\\ \Leftrightarrow5x=35\\ \Leftrightarrow x=7\)
6 tháng 4 2021
\(\dfrac{5x}{1.6}+\dfrac{5x}{6.11}+\dfrac{5x}{11.16}+\dfrac{5x}{16.21}+\dfrac{5x}{21.26}+\dfrac{5x}{26.31}=1\)
\(=x\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+\dfrac{5}{26.31}\right)=1\)
\(=x\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{31}\right)=1\)
\(=x\left(1-\dfrac{1}{31}\right)=1\)
\(\Rightarrow x=1:\left(1-\dfrac{1}{31}\right)=\dfrac{31}{30}\)
đề bài => \(\dfrac{366}{15}x=8x\)=>x=0