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c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
\(\begin{array}{l}a)2x + \frac{1}{2} = \frac{7}{9}\\2x = \frac{7}{9} - \frac{1}{2}\\2x = \frac{{14}}{{18}} - \frac{9}{{18}}\\2x = \frac{5}{{18}}\\x = \frac{5}{{18}}:2\\x = \frac{5}{{18}}.\frac{1}{2}\\x = \frac{5}{{36}}\end{array}\)
Vậy \(x = \frac{5}{{36}}\)
\(\begin{array}{l}b)\frac{3}{4} - 6x = \frac{7}{{13}}\\ 6x = \frac{3}{{4}} - \frac{7}{13}\\ 6x = \frac{{39}}{{52}} - \frac{{28}}{{52}}\\ 6x = \frac{{11}}{{52}}\\x = \frac{{11}}{{52}}:6\\x = \frac{{11}}{{52}}.\frac{{1}}{6}\\x = \frac{{11}}{{312}}\end{array}\)
Vậy \(x = \frac{{11}}{{312}}\)
1/ \(\frac{1}{3x}:\frac{2}{3}=1\)
<=> \(\frac{3}{3×2×x}=\:1\)
<=> \(\frac{1}{2x}=1\)<=> x = \(\frac{1}{2}\)
a) \(-\dfrac{2}{5}+\dfrac{5}{6}x=-\dfrac{4}{15}\\ \Leftrightarrow\dfrac{5}{6}x=\dfrac{2}{15}\\ \Leftrightarrow x=\dfrac{4}{25}\)
b) \(\dfrac{2}{3}+\dfrac{7}{4}\div x=\dfrac{5}{6}\\ \Leftrightarrow\dfrac{7}{4}\div x=\dfrac{1}{6}\\ \Leftrightarrow x=\dfrac{7}{24}\)
a: Ta có: \(-\dfrac{2}{5}+\dfrac{5}{6}x=\dfrac{-4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{5}{6}=\dfrac{2}{15}\)
hay \(x=\dfrac{4}{25}\)
b: Ta có: \(\dfrac{7}{4}:x+\dfrac{2}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{7}{4}:x=\dfrac{1}{6}\)
hay \(x=\dfrac{21}{2}\)
Ta có: \(\left|\frac{5}{6}x+\frac{7}{15}\right|=9-\frac{3}{11}x\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{6}x+\frac{7}{15}=9-\frac{3}{11}x\\\frac{5}{6}x+\frac{7}{15}=\frac{3}{11}x-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{73}{66}x=\frac{128}{15}\Rightarrow x=\frac{2816}{365}\\\frac{37}{66}x=-\frac{142}{15}\Rightarrow x=\frac{-3124}{185}\end{cases}}\)
Vậy x có 2 giá trị như trên