Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 5:\\ a.\ x=\frac{1}{3} \ b.x=5\ hoặc\ x=-1;\\ \ c.\ x=-1;\ d.\ \ x=\frac{8}{15} \ hoặc\ x=\frac{2}{15}\\ e.\ x=\frac{1}{5} ;\ f.\ x=3;\ g.\ x=\frac{1}{2} \ và\ y=-\frac{1}{3} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 5:\\ a.\ \left( x-\frac{1}{3}\right)^{2} =0\Leftrightarrow x-\frac{1}{3} =0\Leftrightarrow x=\frac{1}{3}\\ b.\ ( x-2)^{2} =9\Leftrightarrow x-2=\pm 3\Leftrightarrow x=5\ hoặc\ x=-1\\ c.\ ( 2x-1)^{3} =-27\Leftrightarrow 2x-1=-3\Leftrightarrow x=-1\\ d.\ \left( x-\frac{1}{3}\right)^{2} =\frac{1}{25} \Leftrightarrow x-\frac{1}{3} =\pm \frac{1}{5} \Leftrightarrow x=\frac{8}{15} \ hoặc\ x=\frac{2}{15}\\ e.\ ( 2x-1)^{2} =\frac{-27}{125} =\left(\frac{-3}{5}\right)^{3} \Leftrightarrow 2x-1=\frac{-3}{5} \Leftrightarrow x=\frac{1}{5}\\ f.\ x^{8} =9x^{6} \Leftrightarrow x^{8} :x^{6} =9\Leftrightarrow x^{2} =3^{2} \Leftrightarrow x=3\\ g.\ \left( x-\frac{1}{2}\right)^{2020} +\left( y+\frac{1}{3}\right)^{2022} =0\ ( *)\\ Ta\ có:\ \left( x-\frac{1}{2}\right)^{2020} \geqslant 0\ \forall x\ ;\ \left( y+\frac{1}{3}\right)^{2022} \geqslant 0\ \forall y\\ nên\ ( *) \Leftrightarrow x-\frac{1}{2} =0\ và\ y+\frac{1}{3} =0\\ \Leftrightarrow x=\frac{1}{2} \ và\ y=-\frac{1}{3} \end{array}$