i) \(\left(x-\dfrac{5}{2}\right)^3=\dfrac{-1}{8}\)
\(\text{⇒}\left(x-\dfrac{5}{2}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\text{⇒}x-\dfrac{5}{2}=-\dfrac{1}{2}\)
\(\text{⇒}x=\dfrac{5}{2}-\dfrac{1}{2}\)
\(\text{⇒}x=2\)
j) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\text{⇒}\left(5x+1\right)^2=\left(\dfrac{6}{7}\right)^2\)
TH1:
\(5x+1=\dfrac{6}{7}\)
\(\text{⇒}5x=-\dfrac{1}{7}\)
\(\text{⇒}x=-\dfrac{1}{35}\)
TH2:
\(5x+1=-\dfrac{6}{7}\)
\(\text{⇒}5x=-\dfrac{13}{7}\)
\(\text{⇒}x=-\dfrac{13}{35}\)
k) \(\left(x-\dfrac{3}{2}\right)^2=\dfrac{9}{16}\)
\(\text{⇒}\left(x-\dfrac{3}{2}\right)^2=\left(\dfrac{3}{4}\right)^2\)
TH1:
\(x-\dfrac{3}{2}=\dfrac{3}{4}\)
\(\text{⇒}x=\dfrac{3}{4}+\dfrac{3}{2}\)
\(\text{⇒}x=\dfrac{9}{4}\)
TH2:
\(x-\dfrac{3}{2}=-\dfrac{3}{4}\)
\(\text{⇒}x=-\dfrac{3}{4}+\dfrac{3}{2}\)
\(\text{⇒}x=\dfrac{3}{4}\)
l) \(\left(\dfrac{1}{2}\right)^{2x+1}=\dfrac{1}{8}\)
\(\text{⇒}\left(\dfrac{1}{2}\right)^{2x+1}=\left(\dfrac{1}{2}\right)^3\)
\(\text{⇒}2x+1=3\)
\(\text{⇒}2x=3-1\)
\(\text{⇒}2x=2\)
\(\text{⇒}x=1\)
