Question

Tìm x, biết:

a, ( x - 1/2 ) mũ 3 = 1/27

b, ( x + 1/2 ) mũ 2 = 4/25

Answer 1

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

<=> \(x-\frac{1}{2}=\frac{1}{3}\)

<=> x = \(\frac{5}{6}\)

b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

<=> \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{array}\right.}\)

Vậy...

Answer 2

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

\(\Rightarrow\left(x-\frac{1}{2}\right)=\left(\frac{1}{3}\right)^3\)

\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)

\(\Rightarrow x=\frac{5}{6}\)

b)\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\pm\left(\frac{2}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}\)

\(\Rightarrow\begin{cases}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}\)

 

Answer 3

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

\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)

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

Vậy \(x=\frac{5}{6}\)

b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}=\left(\frac{2}{5}\right)^2=\left(\frac{-2}{5}\right)^2\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=\frac{-2}{5}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-1}{10}\\x=\frac{-9}{10}\end{array}\right.\)

Vậy \(\left[\begin{array}{nghiempt}x=\frac{-1}{10}\\x=\frac{-9}{10}\end{array}\right.\)

Answer 4

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

\(\left(x-\frac{1}{2}\right)^3=\left(\pm\frac{1}{3}\right)^3\)

\(\Rightarrow x-\frac{1}{2}=\pm\frac{1}{3}\)

\(\Rightarrow\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=\frac{-1}{3}\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{1}{3}+\frac{1}{2}\\x=-\frac{1}{3}+\frac{1}{2}\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}\)

Vậy \(x\in\left\{\frac{5}{6};\frac{1}{6}\right\}\)

b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{2}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}\)

\(\Rightarrow\begin{cases}x=\frac{2}{5}-\frac{1}{2}\\x=-\frac{2}{5}-\frac{1}{2}\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}\)

Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}\)