1.
a/ ĐKXĐ: \(-1\le x\le5\)
\(\Leftrightarrow\sqrt{x+3}\le\sqrt{5-x}+\sqrt{x+1}\)
\(\Leftrightarrow x+3\le6+2\sqrt{\left(5-x\right)\left(x+1\right)}\)
\(\Leftrightarrow x-3\le2\sqrt{-x^2+4x+5}\)
- Với \(x< 3\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP\ge0\end{matrix}\right.\) BPT luôn đúng
- Với \(x\ge3\) cả 2 vế ko âm, bình phương:
\(x^2-6x+9\le-4x^2+16x+20\)
\(\Leftrightarrow5x^2-22x-11\le0\) \(\Rightarrow\frac{11-4\sqrt{11}}{5}\le x\le\frac{11+4\sqrt{11}}{5}\)
\(\Rightarrow3\le x\le\frac{11+4\sqrt{11}}{5}\)
Vậy nghiệm của BPT đã cho là \(-1\le x\le\frac{11+4\sqrt{11}}{5}\)
1b/
Đặt \(\sqrt{2x^2+8x+12}=t\ge2\)
\(\Rightarrow x^2+4x=\frac{t^2}{2}-6\)
BPT trở thành:
\(\frac{t^2}{2}-12\ge t\Leftrightarrow t^2-2t-24\ge0\) \(\Rightarrow\left[{}\begin{matrix}t\le-4\left(l\right)\\t\ge6\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+8x+12}\ge6\)
\(\Leftrightarrow2x^2+8x-24\ge0\Rightarrow\left[{}\begin{matrix}x\le-6\\x\ge2\end{matrix}\right.\)
Bài 2:
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\sqrt{x^2+7}-4+\sqrt{x-2}-1=0\)
\(\Leftrightarrow\frac{x^2-9}{\sqrt{x^2+7}+4}+\frac{x-3}{\sqrt{x-2}+1}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{x+3}{\sqrt{x^2+7}+4}+\frac{1}{\sqrt{x-2}+1}\right)=0\)
\(\Rightarrow x=3\)
b/ ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow x^2+x-6+\sqrt{2x-1}-\sqrt{x+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)+\frac{x-2}{\sqrt{2x-1}+\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3+\frac{1}{\sqrt{2x-1}+\sqrt{x+1}}\right)=0\)
\(\Rightarrow x=2\)