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30% . x - x - \(\dfrac{5}{6}\) = \(\dfrac{1}{3}\)
\(\Rightarrow\) (\(\dfrac{3}{10}\) - 1) . x = \(\dfrac{1}{3}+\dfrac{5}{6}\)
\(\Rightarrow\) \(-\dfrac{7}{10}\) . x = \(\dfrac{7}{6}\)
\(\Rightarrow\) x = \(-\dfrac{3}{5}\)
KO ghi đề nhé
\(\dfrac{3}{10}.x-x=\dfrac{1}{3}+\dfrac{5}{6}\)
\(x\left(\dfrac{3}{10}-1\right)=\dfrac{2}{6}+\dfrac{5}{6}\)
\(x.\dfrac{-7}{10}=\dfrac{7}{6}\)
\(x=\dfrac{7}{10}:\dfrac{-7}{6}\)
\(x=\dfrac{7}{10}.\dfrac{6}{-7}\)
\(x=\dfrac{42}{-70}\)
\(x=\dfrac{6}{-10}\)
\(x=\dfrac{3}{-5}\)
Vậy \(x=\dfrac{3}{-5}\)
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f(x)=0 \(\Leftrightarrow\) 2x+a2-3=0 \(\Rightarrow\) x=\(\dfrac{3-a^2}{2}\).
a) x=1 \(\Leftrightarrow\) \(\dfrac{3-a^2}{2}\)=1 \(\Rightarrow\) a=\(\pm\)1.
b) x=\(\dfrac{-1}{2}\) \(\Leftrightarrow\) \(\dfrac{3-a^2}{2}\)=\(\dfrac{-1}{2}\) \(\Rightarrow\) a=\(\pm\)2.
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\(2^x+2^{x+1}+2^{x+2}=960-2^{x+3}\)
\(\Leftrightarrow2^x+2^{x+1}+2^{x+2}+2^{x+3}=960\)
\(\Leftrightarrow2^x\left(1+2+2^2+2^3\right)=960\)
\(\Leftrightarrow2^x.15=960\)
\(\Leftrightarrow2^x=64\)
\(\Leftrightarrow2^x=2^6\Leftrightarrow x=6\)
Vậy...
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\(\Leftrightarrow4x-8+7⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1;9;-5\right\}\)
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a) Có \(\frac{x-1}{9}\)=\(\frac{8}{3}\)
=> \(\left(x-1\right)\times3=8\times9\)
=>\(\left(x-1\right)\times3=27\)
=>\(x-1=27:3\)=>x-1=9=>x=9+1=10
Vậy x=10
b) Có \(\frac{-x}{4}\)=\(\frac{9}{4}\)
Vì 2 phân số này bằng nhau mà có mẫu bằng nhau nên tử cũng phải bằng nhau hay x=-9
c) Có \(\frac{x}{4}=\frac{18}{x+1}\Rightarrow x\left(x+1\right)=18\cdot4=72\)\(\Rightarrow x\times x+x\times1=72\)\(\Rightarrow2x+x=72\Rightarrow3x=72\Rightarrow x=72:3=9\)
Vậy x=9
\(\frac{x+5}{3}-\frac{1}{2}=\frac{1}{3}\)
\(\frac{x+5}{3}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
\(x+5=\frac{5}{6}.3=\frac{5}{2}\)
\(x=\frac{5}{2}-5\)
\(x=\frac{5}{2}-\frac{10}{2}=\frac{-5}{2}\)