a) $x^{2}$ =81
81=$9^{2}$ hoặc $-9^{2}$
=>$x=9;-9$
b) $x^{2}$ =$-100$
$-100$= $-(10)^{2}$
=>$x=-(10)$
c) $(x-3)^{2}$ =$100$
=>$(x-3)^{2}$=$10^{2}$ =$-10^{2}$
=>\(\left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right.\) =>\(\left[ \begin{array}{l}x=13 \\x=-7\end{array} \right.\)
d) $(3x+5)^{2}$ =$49$
=>$(3x+5)^{2}$=$7^{2}$ =$-7^{2}$
=>\(\left[ \begin{array}{l}3x=2\\3x=-12\end{array} \right.\) =>\(\left[ \begin{array}{l}x=\frac{2}{3}\\x=-4\end{array} \right.\)
e) $(3x+5)^{3}$=$729$
=>$(3x+5)^{3}$=$9^{3}$
=>$3x+5=9$
=>$3x=4$
=>$x=\frac{4}{3}$
g) $(5x-2)^{2}$=$100$
=>$(5x-2)^{2}$=$10^{2}$ =$-10^{2}$
=>\(\left[ \begin{array}{l}5x-2=10\\5x-2=-10\end{array} \right.\)
=>\(\left[ \begin{array}{l}5x=12\\5x=-8\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=\frac{12}{5}\\x=\frac{-8}{5}\end{array} \right.\)
h) $(5x+7)^{3}$ =$1000$
=> $(5x+7)^{3}$ =$10^{3}$
=>$x=5x=3$
=>$x=\frac{3}{5}$
k) $(4x-2)^{5}$ =$243$
=>$(4x-2)^{5}$=$3^{5}$
=>$4x-2=3$
=>$4x=5$
=>$x=\frac{5}{4}$
