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

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

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

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

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

Hummmm

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 \(S=R/\left(-3;0\right)\)

ĐKXĐ: \(\begin{cases}3x^2-7x+3\ge0\\ x^2-3x+4\ge0\\ x^2-2\ge0\\ 3x^2-5x-1\ge0\end{cases}\)

=>\(\left[\begin{array}{l}x\le-\sqrt2\\ x\ge\frac{5+\sqrt{37}}{6}\end{array}\right.\)

BPT =>\(\sqrt{3x^2 - 7x + 3} - \sqrt{3x^2 - 5x - 1} > \sqrt{x^2 - 2} - \sqrt{x^2 - 3x + 4}\)

=>\(\dfrac{(3x^2 - 7x + 3) - (3x^2 - 5x - 1)}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}} > \dfrac{(x^2 - 2) - (x^2 - 3x + 4)}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\)

=>\(\dfrac{-2x + 4}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}>\dfrac{3x - 6}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\)

=>\(\dfrac{-2(x - 2)}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}-\dfrac{3(x - 2)}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}>0\)

=>\((x-2)\left[\dfrac{-2}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}-\dfrac{3}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\right]>0\)

=>\((x - 2) \left[ \dfrac{2}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}} + \dfrac{3}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}} \right] < 0\)

=>x-2<0

=>x<2

Kết hợp ĐKXĐ, ta được: \(\left[\begin{array}{l}x\le-\sqrt2\\ \frac{5+\sqrt{37}}{6}\le x<2\end{array}\right.\)

Vậy: \(S = (-\infty, -\sqrt{2}] \cup \left[\dfrac{5+\sqrt{37}}{6}, 2\right)\)

Bạn coi lại đề xem có sai không chứ nghiệm giải ra xấu cực. Và phương trình không rút gọn hết nghe cũng rất vô lý.

dạ vâng,em cx không bt có sai ko do đây là đề của thầy em đưa,chắc cx có sai sót mong thầy bỏ qua