Đáp án:
a.$x\in\{\dfrac{\sqrt{6}+1}{2},-\dfrac{\sqrt{6}+1}{2}\}$
b.$x=49$
c.$0\le x<8$
Giải thích các bước giải:
$a.4x^2+1=8-2\sqrt{6}$
$\rightarrow 4x^2=7-2\sqrt{6}$
$\rightarrow (2x)^2=6-2\sqrt{6}+1$
$\rightarrow (2x)^2=(\sqrt{6}+1)^2$
$\rightarrow 2x=\sqrt{6}+1\rightarrow x=\dfrac{\sqrt{6}+1}{2}$
Hoặc $ 2x=-(\sqrt{6}+1)\rightarrow x=-\dfrac{\sqrt{6}+1}{2}$
$b.2\sqrt{x}=14$
$\rightarrow \sqrt{x}=7$
$\rightarrow x=7^2$
$\rightarrow x=49$
c.ĐKXĐ: $x\ge 0$
$\sqrt{2x}<4$
$\rightarrow 2x<16$
$\rightarrow x<8$
$\rightarrow 0\le x<8$
