ĐK: \(-1\le x\le4\)

\(\sqrt{x+1}+\sqrt{4-x}=t\left(\sqrt{5}\le t\le\sqrt{10}\right)\Rightarrow\sqrt{-x^2+3x+4}=\dfrac{t^2-5}{2}\)

\(pt\Leftrightarrow t+\dfrac{t^2-5}{2}=5\)

\(\Leftrightarrow t^2+2t-15=0\)

\(\Leftrightarrow\left(t-3\right)\left(t+5\right)=0\)

\(\Leftrightarrow t=3\left(\text{Vì }\sqrt{5}\le t\le\sqrt{10}\right)\)

\(\Leftrightarrow\sqrt{x+1}+\sqrt{4-x}=3\)

\(\Leftrightarrow5+2\sqrt{-x^2+3x+4}=9\)

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

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

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

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:

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

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

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

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

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

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

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

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

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

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

ĐKXĐ: 0<=x<=3/2

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

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

Đặt \(u=\sqrt{3x + 2}\ge0;v=\sqrt{3x - 2x^2}\ge0\)

Phương trình sẽ trở thành: \(-3x^2 + x + 3 + uv - 4v + (x - 1)u = 0\)

=>\(v^2-(x-1)(x+3)+uv-4v+(x-1)u=0\)

\(\iff\Big[v^2+uv-(x+3)v-v\Big]+(x-1)u-(x-1)(x+3)=0\)

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

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

TH1: \(x - 1 + \sqrt{3x - 2x^2} = 0\)

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

=>1-x>=0 và \(3x-2x^2=\left(1-x\right)^2\)

=>x<=1 và \(3x-2x^2-x^2+2x-1=0\)

=>x<=1 và \(-3x^2+5x-1=0\)

=>\(x=\frac{5-\sqrt{13}}{6}\)

TH2: \(\sqrt{3x + 2} + \sqrt{3x - 2x^2} - 3x - 1 = 0\)

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

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

=>\(6x+2-2x^2+2\sqrt{-6x^3 + 5x^2 + 6x}=9x^2+6x+1\)

=>\(2\sqrt{-6x^3 + 5x^2 + 6x}=11x^2-1\)

=>\(4(-6x^3 + 5x^2 + 6x) = (11x^2 - 1)^2\) và \(11x^2-1\ge0\)

=>\(-24x^3+20x^2+24x=121x^4-22x^2+1\) và \(11x^2\ge1\)

=>\(121x^4+24x^3-42x^2-24x+1=0\) và \(x^2\ge\frac{1}{11}\)

=>x≃0,6092(nhận)

ĐKXĐ: \(-\dfrac{1}{3}\le x\le4\)

\(\Leftrightarrow x+5=\sqrt{3x+1}+2\sqrt{4-x}\)

Ta có: 

\(VP=1.\sqrt{3x+1}+2.\sqrt{4-x}\le\dfrac{1}{2}\left(1+3x+1\right)+\dfrac{1}{2}\left(4+4-x\right)=x+5\)

\(\Rightarrow VP\le VT\)

Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\sqrt{3x+1}=1\\\sqrt{4-x}=2\end{matrix}\right.\) \(\Leftrightarrow x=0\)