a) (2x+3) ²=9/121
ta có \(\dfrac{9}{121}=\left(\dfrac{3}{11}\right)^2=\left(-\dfrac{3}{11}\right)^2\)
=> (2x+3) ∈ \(\left\{\dfrac{3}{11};\dfrac{-3}{11}\right\}\)
=> x ∈ \(\left\{\dfrac{-15}{11};-\dfrac{18}{11}\right\}\)
Vậy x ∈ \(\left\{\dfrac{-15}{11};-\dfrac{18}{11}\right\}\)
b) (3x-1) ³=-8/27
Ta có \(\dfrac{-8}{27}=\left(\dfrac{-2}{3}\right)^3\)
=> 3x-1 =\(\dfrac{-2}{3}\)
=> x = \(\dfrac{1}{9}\)
Vậy x = \(\dfrac{1}{9}\)
a, \(\left(2x+3\right)^2=\dfrac{9}{121}\Rightarrow\left(2x+3\right)^2=\left(\sqrt{\dfrac{9}{121}}\right)^2\)
\(\Rightarrow2x+3=\sqrt{\dfrac{9}{121}}=\pm\dfrac{3}{11}\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=\dfrac{3}{11}\\2x+3=\dfrac{-3}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\left(\dfrac{3}{11}-3\right)}{2}\\x=\dfrac{\left(\dfrac{-3}{11}-3\right)}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-15}{11}\\x=\dfrac{-18}{11}\end{matrix}\right.\)
b, \(\left(3x-1\right)^3=\dfrac{-8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\dfrac{-8}{27}\Rightarrow\left(3x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(\Rightarrow3x-1=\dfrac{-2}{3}\Rightarrow3x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{9}\)
