b1:
\(-5.\left(x-\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\Leftrightarrow-5x+1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow-\frac{11}{2}x+\frac{4}{3}=\frac{3}{2}x-\frac{5}{6}\Leftrightarrow-\frac{11}{2}x-\frac{3}{2}x=-\frac{5}{6}-\frac{4}{3}\Leftrightarrow-7x=-\frac{13}{6}\)
\(\Leftrightarrow x=\frac{-13}{6}:\left(-7\right)=\frac{13}{42}\)
b2:
a)\(3^{3x-1}=9^{x-2}\Leftrightarrow3^{3x-1}=\left(3^2\right)^{x-2}\Leftrightarrow3^{3x-1}=3^{2x-4}\Leftrightarrow3x-1=2x-4\)
<=>3x-2x=-4+1<=>x=-3
b)\(\left(x-1\right)^4=\left(x-1\right)^6\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^6=0\Leftrightarrow\left(x-1\right)^4\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(x-1\right)^4\left[1-\left(x^2-2x+1\right)\right]=0\Leftrightarrow\left(x-1\right)^4\left(1-x^2+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^4\left(-x^2+2x\right)=0\Leftrightarrow\left(x-1\right)^4.\left(-x\right).\left(x-2\right)=0\)
<=>(x-1)4=0 hoặc -x=0 hoặc x-2=0 <=> x=1 hoặc x=0 hoặc x=2
