Question

tìm x, biết :

a) (2x + 3)² = \(\frac{9}{121}\)

b) (3x - 1)³ = -\(\frac{8}{27}\)

Answer 1

a) (2x + 3)² = 9/121 

Ta có: 9/121 = (3/11)² = (-3/11)²

=> 2x + 3 thuộc {3/11; -3/11}

=> x thuộc {-15/11; -18/11}

b) (3x - 1)³ = -8/27 = (-2/3)³

=> 3x - 1 = -2/3 

=> x = 1/9 

Answer 2

\(\left(2x+3\right)^2=\frac{9}{121}\)

\(\Rightarrow\left(2x+3\right)^2=\hept{\begin{cases}\left(\frac{3}{11}\right)^2\\\left(\frac{-3}{-11}\right)^2\end{cases}}\)

\(\Rightarrow2x+3=\hept{\begin{cases}\frac{3}{11}\\\frac{-3}{-11}\end{cases}}\)

Answer 3

a)\(\left(2x+3\right)^2=\frac{9}{121}\)

\(\Leftrightarrow\left(2x+3\right)^2=\frac{3^3}{11^2}\)

\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{3}{11}\right)^2\)

\(\Leftrightarrow\hept{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=-\frac{30}{11}\Leftrightarrow x=-\frac{15}{11}\\2x=-\frac{36}{11}\Leftrightarrow x=-\frac{18}{11}\end{cases}}}\)

b)\(\left(3x-1\right)^3=-\frac{8}{27}\)

\(\Leftrightarrow\left(3x-1\right)^3=\sqrt[3]{\left(-\frac{8}{27}\right)}\)

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

\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)