\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)

Lời giải:

a) ĐK: $x\geq 0$

BPT $\Leftrightarrow \sqrt{x+2}(\sqrt{2}-1)\leq \sqrt{x}$

$\Leftrightarrow (3-2\sqrt{2})(x+2)\leq x$

$\Leftrightarrow x(2-2\sqrt{2})\leq 2(2\sqrt{2}-3)$

$\Leftrightarrow x\geq \frac{2(2\sqrt{2}-3)}{2-2\sqrt{2}}=-1+\sqrt{2}$

Vậy BPT có nghiệm $x\geq -1+\sqrt{2}$

b) ĐK: $x\geq 2$ hoặc $x\leq -2$

BPT $\Leftrightarrow (x-5)\sqrt{x^2-4}-(x-5)(x+5)\leq 0$

$\Leftrightarrow (x-5)[\sqrt{x^2-4}-(x+5)]\leq 0$Ta có 2 TH:

TH1: 

\(\left\{\begin{matrix} x-5\geq 0\\ \sqrt{x^2-4}-(x+5)\leq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ \sqrt{x^2-4}\leq x+5\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ x^2-4\leq x^2+10x+25\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ 29\leq 10x\end{matrix}\right.\Leftrightarrow x\geq 5\)

TH2: 

\(\left\{\begin{matrix} x-5\leq 0\\ \sqrt{x^2-4}-(x+5)\geq 0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x\leq 5\\ x^2-4\geq x^2+10x+25 \end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 5\\ -29\geq 10x\end{matrix}\right.\)

 \(\Leftrightarrow \left\{\begin{matrix} x\leq 5\\ x\leq \frac{-29}{10}\end{matrix}\right.\Leftrightarrow x\leq \frac{-29}{10}\)

Kết hợp đkxđ suy ra $x\geq 5$ hoặc $x\leq \frac{-29}{10}$

Em 2k8 k biết làm có đúng k

ĐKXĐ : \(\left[{}\begin{matrix}x\le-1\\x\ge3\end{matrix}\right.\) 

Bpt \(\Leftrightarrow\left(x-2\right)\left[x+2-\sqrt{x^2-2x-3}\right]\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge2;x+2\ge\sqrt{x^2-2x-3}\left(1\right)\\x\le2;x+2\le\sqrt{x^2-2x-3}\left(2\right)\end{matrix}\right.\)

(1) có :  \(x+2\ge\sqrt{x^2-2x-3}\Leftrightarrow\left(x+2\right)^2\ge x^2-2x-3\)

\(\Leftrightarrow6x+7\ge0\)  (Đ với \(x\ge2\) )

(2) có : \(\sqrt{x^2-2x-3}\ge x+2\)

TH1 : x + 2 < 0 <=> \(x< -2\)  ( Bpt luôn đúng ) 

TH2 : \(x+2\ge0\) ; Bp 2 vế rút gọn được : \(6x+7\le0\Leftrightarrow x\le\dfrac{-7}{6}\)

Khi đó : \(-2\le x\le\dfrac{-7}{6}\)

Vậy ... 

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

ĐK: \(x\ge2\)

\(\dfrac{\sqrt{x^2+1}-\sqrt{x+1}}{x^2+\sqrt{3x-6}}\ge0\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x^2+1\ge x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-1\le x\le0\\x\ge1\end{matrix}\right.\)

Kết hợp điều kiện xác định ta được \(x\ge2\)

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