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a) \(\frac{3}{4}.x+40\%=\frac{-1}{4}\)
\(\frac{3}{4}.x+\frac{2}{5}=\frac{-1}{4}\)
\(\frac{3}{4}.x=\frac{-13}{20}\)
\(x=\frac{-13}{15}\)
Vậy \(x=\frac{-13}{15}\)
c) \(|x-1|=2^3+\left(-5\right)\)
\(|x-1|=3\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy \(x\in\left\{-2;4\right\}\)
\(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
=>\(\left(x-\frac{1}{3}\right)^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)
=>\(x-\frac{1}{3}=\frac{1}{2}\)hoặc \(x-\frac{1}{3}=-\frac{1}{2}\)
=>x=1/2+1/3hoawcj x=-1/2+1/3
=>x=5/6 hoặc x=-1/6
c)\(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)
\(\Rightarrow x.\left(\frac{1}{2}-\frac{1}{8}\right)=\frac{3}{4}\)
\(\Rightarrow x.\frac{3}{8}=\frac{3}{4}\)
=>x\(=\frac{3}{4}:\frac{3}{8}\)
=>x=\(2\)
a)\(x+\frac{1}{6}=\frac{-3}{8}\)
=>\(x=\frac{-3}{8}-\frac{1}{6}\)
=>\(x=\frac{-9}{24}-\frac{4}{24}\)
=>\(x=\frac{-13}{24}\)
b)\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=2-\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=\frac{24}{12}-\frac{7}{12}\)
\(\Rightarrow\left|\frac{3}{4}-x\right|=\frac{17}{12}\)
TH1: \(\frac{3}{4}-x=\frac{17}{12}\)
=>x=\(\frac{3}{4}-\frac{17}{12}\)
=>x=\(x=-\frac{2}{3}\)
TH2:\(\frac{3}{4}-x=-\frac{17}{12}\)
=>\(x=\frac{3}{4}-\left(-\frac{17}{12}\right)\)
=>x=\(x=\frac{13}{6}\)
Dzồi nhìu phết
\(x+y+y+z+z+x=2\left(x+y+z\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\)
\(\Rightarrow x+y+z=\frac{13}{24}\Rightarrow x=\frac{5}{24};y=\frac{7}{24};z=\frac{1}{24}\)
vậy \(\left(x;y;z\right)=\left(\frac{5}{24};\frac{7}{24};\frac{1}{24}\right)\)
x + y + z =( 1/2 + 1/3 + 1/4 ) : 2
= 13/24
từ đó tính x y z
3-(3/4+x-1/3):2/3=1/2
3-(3/4+x-1/3)=1/2*2/3
3-(3/4+x-1/3)=2/6
3/4+x-1/3=3-2/6
3/4+x-1/3=8/3
3/4+x=8/3+1/3
3/4+x=3
x=3-3/4
x=9/4