\(x\ge9\Rightarrow x+9\ge18\Rightarrow\sqrt{x+9}\ge3\sqrt{2}\)

nguyễn thị thanh huyền

b/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge-\frac{2}{3}\\x\le-1\end{matrix}\right.\)

Đặt \(3x^2+5x+2=t\ge0\)

\(\Leftrightarrow\sqrt{t+5}-\sqrt{t}>1\)

\(\Leftrightarrow\sqrt{t+5}>\sqrt{t}+1\)

\(\Leftrightarrow t+5>t+1+2\sqrt{t}\)

\(\Leftrightarrow\sqrt{t}< 2\Rightarrow t< 4\)

\(\Rightarrow3x^2+5x+2< 4\)

\(\Leftrightarrow3x^2+5x-2< 0\) \(\Rightarrow-2< x< \frac{1}{3}\)

Kết hợp ĐKXĐ ta được nghiệm của BPT:

\(\left[{}\begin{matrix}-2< x\le-1\\-\frac{2}{3}\le x< \frac{1}{3}\end{matrix}\right.\)

e/

ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow x^2+8x-2+6\sqrt{x\left(x+1\right)\left(x-2\right)}\le5x^2-4x-6\)

\(\Leftrightarrow3\sqrt{x\left(x+1\right)\left(x-2\right)}\le2x^2-6x-2\)

\(\Leftrightarrow3\sqrt{\left(x^2-2x\right)\left(x+1\right)}\le2x^2-6x-2\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x}=a\ge0\\\sqrt{x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2-2b^2=2x^2-6x-2\)

BPT trở thành:

\(3ab\le2a^2-2b^2\Leftrightarrow2a^2-3ab-2b^2\ge0\)

\(\Leftrightarrow\left(2a+b\right)\left(a-2b\right)\ge0\)

\(\Leftrightarrow a\ge2b\Rightarrow\sqrt{x^2-2x}\ge2\sqrt{x+1}\)

\(\Leftrightarrow x^2-2x\ge4x+4\)

\(\Leftrightarrow x^2-6x-4\ge0\)

\(\Rightarrow x\ge3+\sqrt{13}\)

d/

ĐKXĐ: \(x\ge-1\)

\(3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}+4x^2-5x+3\ge0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow4a^2-b^2=4x^2-5x+3\)

BPT trở thành:

\(4a^2+3ab-b^2\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(4a-b\right)\ge0\)

\(\Leftrightarrow4a-b\ge0\Rightarrow4a\ge b\)

\(\Rightarrow4\sqrt{x^2+x+1}\ge\sqrt{x+1}\)

\(\Leftrightarrow16x^2+16x+4\ge x+1\)

\(\Leftrightarrow16x^2+15x+3\ge0\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le\frac{-15-\sqrt{33}}{32}\\x\ge\frac{-15+\sqrt{33}}{32}\end{matrix}\right.\)

1/ \(3x^2+4x-3=4x\sqrt{4x-3}\)

\(\Leftrightarrow\left(4x^2-4x\sqrt{4x-3}+4x-3\right)-x^2=0\)

\(\Leftrightarrow\left(2x-\sqrt{4x-3}\right)^2-x^2=0\)

\(\Leftrightarrow\left(3x-\sqrt{4x-3}\right)\left(x-\sqrt{4x-3}\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}3x=\sqrt{4x-3}\\x=\sqrt{4x-3}\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=1\\x=3\end{matrix}\right.\)

3.\(pt\Leftrightarrow\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)

\(\Leftrightarrow\frac{3x+8-5x+4}{\sqrt{3x+8}+\sqrt{5x+4}}-\frac{3x+5-5x+7}{\sqrt{3x+5}+\sqrt{5x+7}}=0\)

\(\Leftrightarrow\left(12-2x\right)\left(\frac{1}{\sqrt{3x+8}+\sqrt{5x+4}}+\frac{1}{\sqrt{3x+5}+\sqrt{5x+7}}\right)=0\)

\(\Rightarrow x=6\)