e) \(\dfrac{1}{2}x+2\dfrac{1}{2}=3\dfrac{1}{2}x-\dfrac{3}{4}\)
\(\text{⇒}\dfrac{1}{2}x+\dfrac{5}{2}=\dfrac{7}{2}x-\dfrac{3}{4}\)
\(\text{⇒}\dfrac{7}{2}x-\dfrac{1}{2}x=\dfrac{5}{2}+\dfrac{3}{4}\)
\(\text{⇒}3x=\dfrac{13}{4}\)
\(\text{⇒}x=\dfrac{13}{4}:3\)
\(\text{⇒}x=\dfrac{13}{12}\)
f) \(2x\left(x-\dfrac{1}{7}\right)=0\)
TH1:
\(2x=0\)
\(\text{⇒}x=0\)
TH2:
\(x-\dfrac{1}{7}=0\)
\(\text{⇒}x=\dfrac{1}{7}\)
g) \(\left(\dfrac{2x}{5}-1\right):\left(-5\right)=\dfrac{1}{4}\)
\(\text{⇒}\dfrac{2x}{5}-1=\dfrac{1}{4}\cdot-5\)
\(\text{⇒}\dfrac{2x}{5}-1=-\dfrac{5}{4}\)
\(\text{⇒}\dfrac{2x}{5}=-\dfrac{5}{4}+1\)
\(\text{⇒}\dfrac{2x}{5}=-\dfrac{1}{4}\)
\(\text{⇒}8x=-5\)
\(\text{⇒}x=-\dfrac{5}{8}\)
h) \(\left(x-1\right)^3=\dfrac{1}{8}\)
\(\text{⇒}\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)
\(\text{⇒}x-1=\dfrac{1}{2}\)
\(\text{⇒}x=\dfrac{1}{2}+1\)
\(\text{⇒}x=\dfrac{3}{2}\)
