a/
ĐKXĐ: \(x\ge\frac{1}{4}\)
\(\Leftrightarrow\sqrt{5x+1}\le\sqrt{4x-1}+3\sqrt{x}\)
\(\Leftrightarrow5x+1\le13x-1+6\sqrt{x\left(4x-1\right)}\)
\(\Leftrightarrow3\sqrt{x\left(4x-1\right)}\ge1-4x\)
Do \(x\ge\frac{1}{4}\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP\le0\end{matrix}\right.\) BPT luôn đúng
Vậy nghiệm của BPT đã cho là \(x\ge\frac{1}{4}\)
b/
ĐKXĐ: \(\left[{}\begin{matrix}x\ge\frac{-5+2\sqrt{5}}{5}\\x\le\frac{-5-2\sqrt{5}}{5}\end{matrix}\right.\)
Đặt \(\sqrt{5x^2+10x+1}=t\ge0\Rightarrow x^2+2x=\frac{t^2-1}{5}\)
BPT trở thành:
\(t\ge7-\frac{t^2-1}{5}\Leftrightarrow t^2+5t-36\ge0\)
\(\Rightarrow\left[{}\begin{matrix}t\le-9\left(l\right)\\t\ge4\end{matrix}\right.\)
\(\Rightarrow\sqrt{5x^2+10x+1}\ge4\)
\(\Leftrightarrow5x^2+10x-15\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)
c/
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow x^2-4+1-\sqrt{x-1}+2-\sqrt{2x}< 0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\frac{x-2}{1+\sqrt{x-1}}-\frac{2\left(x-2\right)}{2+\sqrt{2x}}< 0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-\frac{1}{1+\sqrt{x-1}}-\frac{2}{2+\sqrt{2x}}\right)< 0\)
\(\Leftrightarrow\left(x-2\right)\left(x+\frac{\sqrt{x+1}}{1+\sqrt{x-1}}+\frac{\sqrt{2x}}{2+\sqrt{2x}}\right)< 0\)
\(\Leftrightarrow x-2< 0\Rightarrow x< 2\) (phần trong ngoặc to luôn dương)
Vậy nghiệm của BPT là \(1\le x< 2\)
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:
\(4a^2+3ab-b^2\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(4a-b\right)\ge0\)
\(\Leftrightarrow4a-b\ge0\Rightarrow4a\ge b\)
\(\Rightarrow4\sqrt{x^2+x+1}\ge\sqrt{x+1}\)
\(\Leftrightarrow16x^2+16x+4\ge x+1\)
\(\Leftrightarrow16x^2+15x+3\ge0\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le\frac{-15-\sqrt{33}}{32}\\x\ge\frac{-15+\sqrt{33}}{32}\end{matrix}\right.\)
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:
\(3ab\le2a^2-2b^2\Leftrightarrow2a^2-3ab-2b^2\ge0\)
\(\Leftrightarrow\left(2a+b\right)\left(a-2b\right)\ge0\)
\(\Leftrightarrow a\ge2b\Rightarrow\sqrt{x^2-2x}\ge2\sqrt{x+1}\)
\(\Leftrightarrow x^2-2x\ge4x+4\)
\(\Leftrightarrow x^2-6x-4\ge0\)
\(\Rightarrow x\ge3+\sqrt{13}\)