a, \(\left(2x^3+3\right)^2=\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)
Nếu \(2x+3=\frac{3}{11}\Rightarrow x=-\frac{15}{11}\)
Nếu \(2x+3=-\frac{3}{11}\Rightarrow x=-\frac{18}{11}\)
b,\(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow x=\frac{1}{9}\)
a, (2x+3)^2 = 9/121
=> 2x+3 = \(\sqrt{\frac{9}{121}}\)= \(\frac{3}{11}\)
=>x= \(\frac{\frac{3}{11}-3}{2}\) = \(-\frac{15}{11}\)
b,(3x-1)\(^3\)= \(-\frac{8}{27}\)
=> \(3x-1=\sqrt[3]{-\frac{8}{27}}=-\frac{2}{3}\)
=>\(x=\frac{-\frac{2}{3}+1}{3}=\frac{1}{9}\)
t còn thiếu 1 nghiệm nữa là \(-\frac{18}{11}\)
