Tìm x, biết: a) $x^{2}$ = 81 b) $x^{2}$ = -100 c) ($x-3)^{2}$ = 100 d) ($3x+5)^{2}$ = 49 e) ($3x+5)^{3}$ = 729 g) ($5x-2)^{2}$ = 100 h) ($5x+7)^{3}

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1. 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) $(4x-2)^{5}$ =$243$

   =>$(4x-2)^{5}$=$3^{5}$ 

   =>$4x-2=3$

   =>$4x=5$

   =>$x=\frac{5}{4}$ 

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2. Đáp án:

    

   Giải thích các bước giải:

   $a, x^{2}=81$ 

   ⇒$x^{2}=9^2$

   ⇒$x^{}=±9$ 

   $b, x^{2}=-100$

   $vì$ $x^{2}$ $\geq0$  $∀x$

   $mà$ $-100<0^{}$ 

   ⇒$ko$ $có$ $x$ $thỏa$ $mãn$

   $c,(x-3)^{2}=100$

   ⇔\(\left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\) 

   $d, (3x+5)^{2}=49$

   ⇔\(\left[ \begin{array}{l}3x+5=7\\3x+5=-7\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=\frac{2}3\\x=-4\end{array} \right.\) 

   $e,(3x+5)^{3}=729$

   ⇔$3x+5^{}=9$ 

   ⇔$3x=^{}4$ 

   ⇔$x^{}=$ $\frac{4}{3}$ 

   $g,(5x-2)^{2}=100$

   ⇔\(\left[ \begin{array}{l}5x-2=10\\5x-2=-10\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=\frac{12}{5}\\x=\frac{-8}5\end{array} \right.\) 

   $h,(5x+7)^{3}=1000$

   ⇔$5x^{}+7=10$ 

   ⇔$5x^{}=3$ 

   ⇔$x=\frac{3}{5}$ 

   $k,(4x-2)^{5}=243$

   ⇔$4x^{}-2=3$ 

   ⇔$4x^{}=5$ 

   ⇔$x=\frac{5}{4}$ 

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