ĐKXĐ: \(x\ge3\)

Khi đó \(\sqrt{2x-1}\ge\sqrt{5}>1\Rightarrow\sqrt{2x-1}-1>0\)

Đồng thời \(\sqrt{x+3}>\sqrt{x-3}\) \(\forall x\Rightarrow\sqrt{x+3}-\sqrt{x-3}>0\)

Do đó BPT tương đương:

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

\(\Leftrightarrow\sqrt{x^2-9}-x+3\ge\sqrt{2x-1}-1\)

\(\Leftrightarrow\sqrt{x^2-9}\ge x-4+\sqrt{2x-1}\)

Do \(x-4+\sqrt{2x-1}\ge3-4+\sqrt{5}>0;\forall x\ge3\) nên BPT tương đương:

\(x^2-9\ge x^2-8x+16+2x-1+2\left(x-4\right)\sqrt{2x-1}\)

\(\Leftrightarrow\left(x-4\right)\sqrt{2x-1}-3\left(x-4\right)\le0\)

\(\Leftrightarrow\left(x-4\right)\left(\sqrt{2x-1}-3\right)\le0\)

\(\Leftrightarrow\left(x-4\right)\left(\frac{2x-1-9}{\sqrt{2x-1}+3}\right)\le0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)\le0\Leftrightarrow4\le x\le5\)

a,ĐK: x\(\ge\)1

⇔\(\sqrt{x-1-2\sqrt{x-1}+1}\)=\(\sqrt{2}\)

⇔\(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)=\(\sqrt{2}\)

⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)

TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1   bn tự giải ra nha

TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\)    bn tự lm nha

\(\sqrt{2x-1}\ge0\)

\(\Rightarrow BPT\ge0\) khi

\(3-2x-x^2\ge0\)

\(\Leftrightarrow x^2+2x-3\le0\)

\(\Leftrightarrow\left(x+1\right)^2-4\le0\)

\(\Leftrightarrow\left(x+1\right)^2\le4\)

\(\Leftrightarrow x+1\le2\)

\(\Rightarrow x\le1\)

ĐKXĐ: \(x^2+x-1\ge0\)

\(\Rightarrow3x^2-x+1>3\sqrt{\left(x^2-x+1\right)\left(x^2+x-1\right)}\)

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

\(\Rightarrow2a^2+b^2>3ab\)

\(\Leftrightarrow\left(2a-b\right)\left(a-b\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}2a< b\\a>b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\sqrt{x^2-x+1}< \sqrt{x^2+x-1}\\\sqrt{x^2-x+1}>\sqrt{x^2+x-1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2-x+1\right)< x^2+x-1\\x^2-x+1>x^2+x-1\end{matrix}\right.\)

\(\Leftrightarrow...\) (nhớ kết hợp ĐKXĐ ban đầu)