Question

tìm x biết
a)
\(\frac{1}{2}\)+\(\frac{2}{3}\).x=\(\frac{4}{5}\)
b)
x+\(\frac{1}{4}\)=\(\frac{4}{3}\)|
c)
\(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
|d)
\(\left|x+5\right|-6=9\)
e)
\(\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
f)
\(\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
g)
\(^{x^2}\)=16
h)
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
i)
\(^{3^3}\).x=\(^{3^6}\)
j)
\(\frac{1,35}{0,2}=\frac{1,25}{x}\)
k)
\(1\frac{2}{3}:x=6:0,3\)

Answer 1

\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)