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a)\(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)
\(2x=3,8:\dfrac{3}{32}\)
\(x=\dfrac{608}{15}:2\)
\(x=\dfrac{304}{15}\)
b)\(\left(0,25x\right):3=\dfrac{5}{6}:0,125\)
\(0,25x=\dfrac{20}{3}\times3\)
\(x=20:0,25\)
\(x=80\)
c) \(0,01:2,5=\left(0,75\times x\right):0,75\)
\(0,004=0,75\times x:0,75\)
\(0,004=x\times0,75:0,75\)
\(0,004=x\times1\)
\(0,004=x\)
\(\Rightarrow x=0,004\)
d)\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:\left(0,1x\right)\)
\(\dfrac{5}{3}=\dfrac{2}{3}:0,1\times x\)
\(\dfrac{5}{3}=\dfrac{20}{3}\times x\)
\(\Rightarrow x=\dfrac{20}{3}:\dfrac{5}{3}\)
\(x=4\)
a) \(x=20\dfrac{4}{15}.\)
b) \(x=80\)
c) \(x=0,004\)
d) \(x=4.\)
a) \(\dfrac{5}{6}:x=30:3\)
\(\Leftrightarrow\dfrac{5}{6}:x=10\)
\(\Leftrightarrow x=\dfrac{5}{6}:10\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy .......
b) \(x:2,5=0,003:0,75\)
\(\Leftrightarrow x:2,5=0,004\)
\(\Leftrightarrow x=0,004.2,5\)
\(\Leftrightarrow x=0,01\)
Vậy .......
c) \(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)
\(\Leftrightarrow3,8:\left(2x\right)=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{3}{32}\)
\(\Leftrightarrow2x=3,8:\dfrac{3}{32}\)
\(\Leftrightarrow2x=\dfrac{698}{25}\)
\(\Leftrightarrow x=\dfrac{304}{15}\)
Vậy ...
d) \(\dfrac{2}{3}:0,4=x:\dfrac{4}{5}\)
\(\Leftrightarrow x:\dfrac{4}{5}=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{8}{15}\)
Vậy ....
e) \(3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)
\(\Leftrightarrow0,25:x=\dfrac{19}{5}:\dfrac{608}{15}\)
\(\Leftrightarrow0,25x=\dfrac{57}{608}\)
\(\Leftrightarrow x=\dfrac{228}{608}\)
Vậy ...
e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow xx=\left(-60\right)\left(-15\right)\)
\(\Leftrightarrow x^2=900\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=30^2\\x^2=\left(-30\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)
Vậy ...
a: =>x/3=-5/2
hay x=-15/2
b: \(\Leftrightarrow\dfrac{7}{3}:x=\dfrac{1}{5}-\dfrac{4}{9}=\dfrac{9-20}{45}=\dfrac{-11}{45}\)
\(\Leftrightarrow x=\dfrac{7}{3}:\dfrac{-11}{45}=\dfrac{7}{3}\cdot\dfrac{-45}{11}=\dfrac{-105}{11}\)
c: \(\Leftrightarrow x=\dfrac{-7}{2}\cdot2=-7\)
d: =>x/27=-1/3+2/9=2/9-3/9=-1/9=-3/27
=>x=-3
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
a) \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
\(\Leftrightarrow3-2x=-1\)
\(\Leftrightarrow-2x=-1-3\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{5}{3}x+\dfrac{1}{5}=x-\dfrac{7}{8}\)
\(\Leftrightarrow200x+24=120x-105\)
\(\Leftrightarrow80x=-129\)
\(\Leftrightarrow x=-\dfrac{129}{80}\)
Vậy \(x=-\dfrac{129}{80}\)
c) \(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
\(\Leftrightarrow-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow-\left|x+0,5\right|=-\dfrac{11}{20}\)
\(\Leftrightarrow\left|x+0,5\right|=\dfrac{11}{20}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+0,5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{21}{20};x_2=\dfrac{1}{20}\)
d) \(\left(x+0,2\right)^2+0,75=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2+\dfrac{3}{4}=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=1-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow x+\dfrac{1}{5}=\pm\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{1}{2}\\x+\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{7}{10};x_2=\dfrac{3}{10}\)
a, \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
=>\(-\dfrac{1}{2}x=-\dfrac{1}{4}-\dfrac{3}{4}\)
=>\(-\dfrac{1}{2}x=-1\)
=>\(x=-1:(-\dfrac{1}{2})\)
=>\(x=2\)
vậy ...........
b,\(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x-x=-\dfrac{7}{8}-0,2\)
=>\(\dfrac{2}{3}x=-\dfrac{43}{40}\)
=>\(x=-\dfrac{43}{40}:\dfrac{2}{3}\)
=>\(x=-\dfrac{192}{80}\)
vậy...................
c,\(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
=>\(-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
=>\(-\left|x+0,5\right|=-\dfrac{11}{20}\)
=>\(\left|x+0,5\right|=\dfrac{11}{20}\)
=>\(\left[{}\begin{matrix}x+0.5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
vậy ....... hoặc.....
d,\((x+0,2)^2+0,75=1\)
=>\(\left(x+0,2\right)^2=1-0,75\)
=>\(\left(x+0,2\right)^2=0,25\)
=>\(\left[{}\begin{matrix}x+0,2=0,5\\x+0,2=-0,5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0,3\\x=-0,7\end{matrix}\right.\)
vậy..........................
HỌC TỐT NHA !!!!!
\(a,\dfrac{x}{6}=\dfrac{7}{3}\Rightarrow x=\dfrac{6.7}{3}\Rightarrow x=14\)
\(b,\dfrac{20}{x}=\dfrac{-12}{15}\Rightarrow x=\dfrac{20.15}{-12}\Rightarrow x=-25\)
\(c,\dfrac{-15}{35}=\dfrac{27}{x}\Rightarrow x=\dfrac{35.27}{-15}\Rightarrow x=-63\)
\(d,\dfrac{\dfrac{4}{5}}{1\dfrac{2}{5}}=\dfrac{2\dfrac{2}{5}}{x}\Rightarrow\dfrac{\dfrac{4}{5}}{\dfrac{7}{5}}=\dfrac{\dfrac{12}{5}}{x}\Rightarrow x=\dfrac{\dfrac{7}{5}.\dfrac{12}{5}}{\dfrac{4}{5}}\Rightarrow x=\dfrac{\dfrac{84}{25}}{\dfrac{4}{5}}\Rightarrow x=\dfrac{21}{5}\)
\(e,\dfrac{x}{1\dfrac{1}{4}}=\dfrac{5}{2}\Rightarrow\dfrac{x}{\dfrac{5}{4}}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{2}.\dfrac{5}{4}\Rightarrow x=\dfrac{25}{8}\)
\(f,\dfrac{\dfrac{1}{2}}{1\dfrac{1}{4}}=\dfrac{x}{3\dfrac{1}{3}}\Rightarrow\dfrac{\dfrac{1}{2}}{\dfrac{5}{4}}=\dfrac{x}{\dfrac{10}{3}}\Rightarrow x=\dfrac{\dfrac{10}{3}.\dfrac{1}{2}}{\dfrac{5}{4}}\Rightarrow x=\dfrac{\dfrac{5}{3}}{\dfrac{5}{4}}\Rightarrow x=\dfrac{4}{3}\)
a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)
Vậy: \(x\in\left\{30;-30\right\}\)
b) Ta có: \(\left|x\right|+0.573=2\)
\(\Leftrightarrow\left|x\right|=1.427\)
hay \(x\in\left\{1.427;-1.427\right\}\)
Vậy: \(x\in\left\{1.427;-1.427\right\}\)
c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)
d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)
\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)
\(\Leftrightarrow x=\dfrac{1}{250}\)
Vậy: \(x=\dfrac{1}{250}\)