Đáp án:
Giải thích các bước giải:
Bài 5:
$a)\sqrt{x^{2}-8x+16}=4 \\\Leftrightarrow \sqrt{(x-4)^{2}}=4\\\Leftrightarrow \left | x-4 \right |=4\\\Leftrightarrow \left[ \begin{array}{l}x-4=4,x-4\geq 0\\-(x-4)=4,x-4<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=8,x\geq 4\\x=0,x<4\end{array} \right.$
Vậy $x\in \left \{ 0;8 \right \}$
$b) \sqrt{-3x+4}=12\ ĐK:x \le \dfrac{4}{3}\\\Leftrightarrow -3x+4=144\\\Leftrightarrow -3x=144-4\\\Leftrightarrow -3x=140\\\Leftrightarrow x=-\frac{140}{3}\ (TM)$
Vậy $x=-\dfrac{140}{3}$
$c) \sqrt{4(1-x)^{2}}-\sqrt{3}=0\\\Leftrightarrow \sqrt{4(1-x)^{2}}=\sqrt{3}\\\Leftrightarrow 4(1-x)^{2}=3\\\Leftrightarrow (1-x)^{2}=\frac{3}{4}\\\Leftrightarrow 1-x=±\dfrac{\sqrt{3}}{2}\\\Leftrightarrow x=±\dfrac{\sqrt{3}}{2}+1$
Vậy $x=±\dfrac{\sqrt{3}}{2}+1$
$d) \sqrt{\sqrt{5}-\sqrt{3}x}=\sqrt{8+2\sqrt{15}}\\\Leftrightarrow \sqrt{5}-\sqrt{3}x=8+2\sqrt{15}\\\Leftrightarrow -\sqrt{3}x=8+2\sqrt{15}-\sqrt{5}\\\Leftrightarrow x=-\dfrac{8}{\sqrt{3}}-2\sqrt{5}+\dfrac{\sqrt{5}}{\sqrt{3}}\\\Leftrightarrow x=-\dfrac{8\sqrt{3}}{3}-2\sqrt{5}+\dfrac{\sqrt{15}}{3}\\\Leftrightarrow x=\dfrac{-8\sqrt{3}+\sqrt{15}}{3}-2\sqrt{5}$
Vậy $x=\dfrac{-8\sqrt{3}+\sqrt{15}}{3}-2\sqrt{5}$
