a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)

\(\Leftrightarrow x^2-2x-3=1\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

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

Lời giải:

Đặt $\sqrt[3]{x^2+3x-5}=a; \sqrt[3]{x+2}=b$. Khi đó pt đã cho tương đương với:

$a+b=\sqrt[3]{a^3+b^3-1}+1$

$\Leftrightarrow a+b-1=\sqrt[3]{a^3+b^3-1}$

$\Leftrightarrow (a+b-1)^3=a^3+b^3-1$

$\Leftrightarrow (a+b)^3-3(a+b)^2+3(a+b)-1=a^3+b^3-1$

$\Leftrightarrow 3ab(a+b)-3(a+b)^2+3(a+b)=0$

$\Leftrightarrow ab(a+b)-(a+b)^2+(a+b)=0$

$\Leftrightarrow (a+b)(ab-a-b+1)=0$

$\Leftrightarrow (a+b)(a-1)(b-1)=0$

Nếu $a+b=0\Leftrightarrow \sqrt[3]{x^2+3x-5}=-\sqrt[3]{x+2}$

$\Leftrightarrow x^2+3x-5=-(x+2)$

$\Leftrightarrow x^2+4x-3=0$

$\Leftrightarrow x=-2\pm \sqrt{7}$

Nếu $a-1=0\Leftrightarrow \sqrt[3]{x^2+3x-5}=1$

$\Leftrightarrow x^2+3x-6=0$

$\Leftrightarrow x=\frac{-3\pm \sqrt{33}}{2}$

Nếu $b-1=0\Leftrightarrow \sqrt[3]{x+2}=1$

$\Leftrightarrow x=-1$

1.

ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)

\(\Leftrightarrow x^2+x-1+2\sqrt{x\left(x^2-1\right)}=2x^2-3x-4\)

\(\Leftrightarrow x^2-4x-3-2\sqrt{\left(x^2-x\right)\left(x+1\right)}=0\)

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

\(\Rightarrow a^2-3b^2-2ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)

\(\Leftrightarrow a=3b\)

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

\(\Leftrightarrow x^2-x=9\left(x+1\right)\)

\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)

2.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:

\(x^3+3\left(x^2-4a^2\right)a=0\)

\(\Leftrightarrow x^3+3ax^2-4a^3=0\)

\(\Leftrightarrow\left(x-a\right)\left(x+2a\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=x\\2a=-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-4x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=2-2\sqrt{2}\end{matrix}\right.\)

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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