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a) \(-\dfrac{3}{5}-x=-0,75\)
\(\Rightarrow-\dfrac{3}{5}-x=-\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{15}{20}-\dfrac{12}{20}=\dfrac{8}{20}=\dfrac{2}{5}\)
b) \(1\dfrac{4}{5}=-0,15-x\)
\(\Rightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{9}{5}\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{36}{20}=-\dfrac{39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{5}\)
a) \(-\dfrac{3}{5}-x=-0,75\)
\(x=-\dfrac{3}{5}+0,75=\dfrac{3}{5}+\dfrac{3}{4}\)
\(x=\dfrac{27}{20}\)
________
b) \(1\dfrac{4}{5}=-0,15-x\)
\(=>-0,15-x=\dfrac{9}{5}\)
\(x=\dfrac{-3}{20}-\dfrac{9}{5}=\dfrac{-3}{20}-\dfrac{36}{20}\)
\(x=\dfrac{-39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)=\dfrac{6}{15}+\dfrac{5}{15}\)
\(x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(x=\dfrac{11}{15}-\dfrac{1}{3}=\dfrac{11}{15}-\dfrac{5}{15}\)
\(x=\dfrac{6}{15}=\dfrac{2}{5}\)
a)Ta có:\(-\frac{2}{7}=\frac{-2.11}{7.11}=-\frac{22}{77}\)
\(-\frac{3}{11}=\frac{-3.7}{11.7}=-\frac{21}{77}\)
Vì \(-\frac{22}{77}< -\frac{21}{77}\)
Vậy \(-\frac{2}{7}< -\frac{3}{11}\)
b)Ta có:\(-\frac{28}{25}=\frac{-28.12}{25.12}=-\frac{336}{300}\)
Vì \(-\frac{336}{300}< -\frac{213}{300}\)
Vậy \(-\frac{28}{25}< -\frac{213}{300}\)
c)Ta có:\(-\frac{3}{4}=-0,75\)
Vì -0,75=-0,75
Vậy \(-\frac{3}{4}=-0,75\)
Lời giải:
$\frac{1}{4}-3x+\frac{3}{2}=-0,75$
$3x=\frac{1}{4}+\frac{3}{2}-(-0,75)=2,5$
$\Rightarrow x=2,5:3=\frac{5}{6}$
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
a) Ta có: \(x-\frac{2}{3}=0,75\)
\(\Leftrightarrow x-\frac{2}{3}=\frac{3}{4}\)
hay \(x=\frac{3}{4}+\frac{2}{3}=\frac{17}{12}\)
Vậy: \(x=\frac{17}{12}\)
b) Ta có: \(\frac{1}{3}+\frac{2}{3}x=-1\)
\(\Leftrightarrow\frac{2}{3}x=-1-\frac{1}{3}=-\frac{4}{3}\)
hay \(x=\frac{-4}{3}:\frac{2}{3}=\frac{-4}{3}\cdot\frac{3}{2}=-2\)
Vậy: x=-2
c) Ta có: \(\left|2x-3\right|-\frac{3}{4}=4,25\)
\(\Leftrightarrow\left|2x-3\right|=\frac{17}{4}+\frac{3}{4}=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy: x∈{-1;4}
a, \(x-\frac{2}{3}=0.75\)
\(x=0.75-\frac{2}{3}\)
\(x=\frac{1}{12}\)
Vậy...
b, \(\frac{1}{3}+\frac{2}{3}\cdot x=-1\)
\(\frac{2}{3}\cdot x=-1-\frac{1}{3}=-\frac{4}{3}\)
\(x=-\frac{4}{3}:\frac{2}{3}=-2\)
Vậy x = -2
c, \(|2x-3|-\frac{3}{4}=4.25\)
\(|2x-3|=4.25-\frac{3}{4}=\frac{7}{2}\)
=> \(2x-3=\frac{7}{2}hay2x-3=-\frac{7}{2}\)
2x = \(\frac{7}{2}+3\) 2x = \(-\frac{7}{2}+3\)
2x = \(\frac{13}{2}\) 2x = \(-\frac{1}{2}\)
x = \(\frac{12}{2}:2=\frac{13}{4}\) x = \(-\frac{1}{2}:2\) = \(-\frac{1}{4}\)
Vậy...
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: \(x=\dfrac{1}{250}\)
a, \(\left|x\right|+0,75=2\Rightarrow\left|x\right|=1,25\Rightarrow x=\pm1,25\)
b, \(\left|x+\dfrac{1}{3}\right|-4=1\Rightarrow\left|x+\dfrac{1}{3}\right|=5\Rightarrow\left|x\right|=\dfrac{14}{3}\Rightarrow x=\pm\dfrac{14}{3}\)
c,
\(\left|x\right|+2,5=0\Rightarrow\left|x\right|=-0,25\) (vô lí)
Vì \(\left|x\right|\ge0\) => x vô nghiệm
a) \(\left|x\right|+0,75=2\)
\(\Leftrightarrow\left|x\right|=2-0,75=1,25\)
\(\Rightarrow\left[{}\begin{matrix}x=-1,25\\x=1,25\end{matrix}\right.\)
b) \(\left|x+\dfrac{1}{3}\right|-4=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=1+4=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=5\\x+\dfrac{1}{3}=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=-\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left|x\right|+2,5=0\)
\(\Leftrightarrow\left|x\right|=0-2,5=-2,5\)
Vì \(\left|x\right|\ge0\) mà \(-2,5< 0\) nên \(x\in\varnothing\)
= 3/4 x 15,2/3 x 3/4 x 24,1/3
=3 x 15,2 x 3 x 24,1 / 4 x 4 x 3 x 3
= 15,2 x 24,1 / 16
=4579/200