Đáp án: a.$x=-1$
b.$x=3$
Giải thích các bước giải:
a.Ta có:
$\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2$
$\to \sqrt{3(x^2+2x+1)+4}+\sqrt{5(x^2+2x+1)+9}=5-(x^2+2x+1)$
$\to \sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}=5-(x+1)^2$
Mà $\sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}\ge \sqrt{3\cdot 0^2+4}+\sqrt{5\cdot 0+9}=5$
$5-(x+1)^2\le 5$
$\to \sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}\ge 5\ge 5-(x+1)^2$
Dấu = xảy ra khi $x+1=0\to x=-1$
b.ĐKXĐ: $2\le x\le 4$
Ta có: $\sqrt{x-2}+\sqrt{4-x}\le \sqrt{2(x-2+4-x)}=2$
Mà $x^2-6x+11=(x^2-6x+9)+2=(x-3)^2+2\ge 2$
$\to \sqrt{x-2}+\sqrt{4-x}\le2\le x^2-6x+11$
$\to$Dấu = xảy ra khi $\begin{cases}\sqrt{x-2}=\sqrt{4-x}\\ x-3=0\end{cases}\to x =3$
