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)

ĐKXĐ: \(\frac{2}{3}\le x\le5\)

\(\Leftrightarrow\sqrt{2x+7}\ge\sqrt{5-x}+\sqrt{3x-2}\)

\(\Leftrightarrow2x+7\ge2x+3+2\sqrt{-3x^2+17x-10}\)

\(\Leftrightarrow\sqrt{-3x^2+17x-10}\le2\)

\(\Leftrightarrow-3x^2+17x-10\le4\)

\(\Leftrightarrow3x^2-17x+14\ge0\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge\frac{14}{3}\end{matrix}\right.\)

Kết hợp ĐKXĐ: \(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}\le x\le1\\\frac{14}{3}\le x\le5\end{matrix}\right.\)

1. Đợi chút t tìm cách ngắn gọn.

2. ĐK: \(\left\{{}\begin{matrix}2x^2+8x+6\ge0\\x^2-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\le-3\\x\ge1\\x=-1\end{matrix}\right.\) (*)

BPT\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\3x^2+8x+5+2\sqrt{\left(2x^2+8x+6\right)\left(x^2-1\right)}\le\left(2x+2\right)^2\left(1\right)\end{matrix}\right.\)

Giải (1) \(\Leftrightarrow x^2-1-2\sqrt{\left(2x^2+8x+6\right)\left(x^2-1\right)}\ge0\)

\(\Leftrightarrow\sqrt{x^2-1}\left(\sqrt{x^2-1}-2\sqrt{2x^2+8x+6}\right)\ge0\)

TH1: \(\sqrt{x^2-1}=0\Leftrightarrow x=\pm1\) (tm)

TH2: \(x^2-1\ne0\)

\(\Leftrightarrow\sqrt{x^2-1}-2\sqrt{2x^2+8x+6}\ge0\)

\(\Leftrightarrow\sqrt{x^2-1}\ge2\sqrt{2x^2+8x+6}\)

\(\Leftrightarrow x^2-1\ge8x^2+32x+24\)

\(\Leftrightarrow7x^2+32x+25\le0\)

\(\Leftrightarrow-\frac{25}{7}\le x\le-1\) kết hợp đk (*) và đk để giải bpt

=>\(x=-1\)

Vậy \(x=\pm1\)

3. ĐK: \(x\ge\frac{4}{5}\)

\(BPT\Leftrightarrow\sqrt{5x-4}-\sqrt{3x-2}+\sqrt{4x-3}-\sqrt{2x-1}>0\)

\(\Leftrightarrow\frac{2x-2}{\sqrt{5x-4}+\sqrt{3x-2}}+\frac{2x-2}{\sqrt{4x-3}+\sqrt{2x-1}}>0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{5x-4}+\sqrt{3x-2}}+\frac{1}{\sqrt{4x-3}+\sqrt{2x-1}}\right)>0\)

\(\Leftrightarrow x-1>0\) \(\Leftrightarrow x>1\)

Vậy \(x>1\)

\(-4x^2-3x^2\ge0\)

\(\Leftrightarrow-x^2\left(2x^2+3\right)\ge0\)

Vì \(-x^2\le0\Rightarrow-x^2\left(2x^2+3\right)\ge0\Leftrightarrow2x^2+3\le0\)

\(\Leftrightarrow2x^2\le-3\)

\(\Leftrightarrow x^2\le\frac{-3}{2}\)(vô lí)

Vậy \(x\in\phi\)

ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge\frac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x-1>\sqrt{2x^2-3x-5}\)

- Với \(x\le-1\Rightarrow\left\{{}\begin{matrix}VT< 0\\VP\ge0\end{matrix}\right.\) BPT vô nghiệm

- Với \(x\ge\frac{5}{2}\) hai vế ko âm, bình phương:

\(x^2-2x+1>2x^2-3x-5\)

\(\Leftrightarrow x^2-x-6< 0\Rightarrow-2< x< 3\)

\(\Rightarrow\frac{5}{2}\le x< 3\)

từ câu 1 đến câu 4 bạn có thẻ dùng máy tính casio f(x)570 VN giải nhé .bạn bấm MODE xuống 1 1

1)vô nghiệm

2)vô nghiệm

3)luôn đúng

4)\(\frac{-1-\sqrt{41}}{4}\le x\le\frac{-1+\sqrt{41}}{4}\)

5) \(\left\{{}\begin{matrix}-2x^2+5x-2\le x-3\\-2x^2+5x-2\ge-x+3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le\frac{2-\sqrt{6}}{2}\\x\ge\frac{2+\sqrt{6}}{2}\end{matrix}\right.\\vonghiem\end{matrix}\right.\) vậy bpt vô nghiệm