Question

Giải các bất phương trình chứa căn sau:

1. \(\sqrt{x+8}\left(\sqrt{x+3}-\sqrt{x}\right)\ge3\)

2. \(5\sqrt{x}+\frac{5}{2\sqrt{x}}< 2x+\frac{1}{2x}+4\)

3. \(\sqrt{x+3}+\sqrt{x+2}>\sqrt{2x+4}\)

HY VỌNG MỌI NGƯỜI GIÚP MÌNH NHA! CẢM ƠN RẤT NHIỀU!

Answer 1

a/ ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow3\sqrt{x+8}\ge3\left(\sqrt{x+3}+\sqrt{x}\right)\)

\(\Leftrightarrow\sqrt{x+8}\ge\sqrt{x+3}+\sqrt{x}\)

\(\Leftrightarrow x+8\ge2x+3+2\sqrt{x^2+3x}\)

\(\Leftrightarrow5-x\ge2\sqrt{x^2+3x}\)

- Với \(x>5\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP\ge0\end{matrix}\right.\) BPT vô nghiệm

- Với \(x\le5\) hai vế ko âm, bình phương:

\(x^2-10x+25\ge4x^2+12x\)

\(\Leftrightarrow3x^2+22x-25\le0\Rightarrow-\frac{25}{3}\le x\le1\)

Vậy nghiệm của BPT đã cho là \(0\le x\le1\)

Answer 2

b/ ĐKXĐ: \(x>0\)

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)< 2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\Rightarrow x+\frac{1}{4x}=t^2-1\)

BPT trở thành:

\(5t< 2\left(t^2-1\right)+4\)

\(\Leftrightarrow2t^2-5t+2>0\Rightarrow\left[{}\begin{matrix}t>2\\t< \frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0\le x< \frac{3-2\sqrt{2}}{2}\\x>\frac{3+2\sqrt{2}}{2}\end{matrix}\right.\)

Answer 3

c/ ĐKXĐ: \(x\ge-2\)

\(\Leftrightarrow\sqrt{x+3}>\sqrt{2\left(x+2\right)}-\sqrt{x+2}\)

\(\Leftrightarrow\sqrt{x+3}>\left(\sqrt{2}-1\right)\sqrt{x+2}\)

\(\Leftrightarrow x+3>\left(3-2\sqrt{2}\right)\left(x+2\right)\)

\(\Leftrightarrow\left(2\sqrt{2}-2\right)x>3-4\sqrt{2}\)

\(\Rightarrow x>\frac{3-4\sqrt{2}}{2\sqrt{2}-2}\)