Đk: x \(\le\)2028

Ta có: \(\sqrt{2028-x}+\sqrt{2093-x}+\sqrt{2268-x}=29\)

<=> \(\sqrt{2028-x}-4+\sqrt{2093-x}-9+\sqrt{2268-x}-16=0\)

<=> \(\frac{2028-x-16}{\sqrt{2028-x}+4}+\frac{2093-x-81}{\sqrt{2093-x}+9}+\frac{2268-x-256}{\sqrt{2268-x}+16}=0\)

<=> \(\left(2012-x\right).\left(\frac{1}{\sqrt{2028-x}+4}+\frac{1}{\sqrt{2093-x}+9}+\frac{1}{\sqrt{2268-x}+16}\right)=0\)

<=> x = 2012 (tm)

\(4\sqrt{x+3}-\sqrt{x-1}=7\left(1\right)\left(ĐK:x\ge1\right)\)

\(\Leftrightarrow4\sqrt{x+3}=7+\sqrt{x-1}\)

\(\Leftrightarrow16\left(x+3\right)=49+x-1+14\sqrt{x+1}\)

\(\Leftrightarrow15x=14\sqrt{x-1}\)\(\left(x\ge1\right)\)

\(\Leftrightarrow225x^2=196\left(x-1\right)\)

\(\Leftrightarrow225x^2-196x+196=0\)

\(\Delta=196^2-4.225.196< 0\)

\(\Rightarrow pt\)vô nghiệm

Vậy pt vô nghiệm.

\(9x^2-\left(3x+2\right)\sqrt{3x-1}+2=3x\left(1\right)\)\(\left(x\ge\frac{1}{3}\right)\)

Đặt \(\sqrt{3x-1}=a\ge0\)

\(\Rightarrow\hept{\begin{cases}3x=a^2+1\\3x+2=a^2+3\\3x-1=a^2\end{cases}}\)

Pt (1) \(\Leftrightarrow3x\left(3x-1\right)-\left(3x+2\right)\sqrt{3x-1}+2=0\)

\(\Leftrightarrow\left(a^2+1\right)a^2-\left(a^2+3\right)a+2=0\)

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

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

\(\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=2\end{cases}}\)

TH1: a=1 

\(\Rightarrow\sqrt{3x-1}=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow x=\frac{2}{3}\left(tm\right)\)

TH2: a=2 

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

\(\Leftrightarrow3x-1=4\)

\(\Leftrightarrow x=\frac{5}{3}\left(tm\right)\)

Vậy pt có tập nghiệm \(S=\left\{\frac{2}{3};\frac{5}{3}\right\}\)

ĐK: \(x\ge1\).

Đặt \(\sqrt{x-1}=a,2x-5=b\).

Phương trình ban đầu tương đương với: 

\(\sqrt{b^2+5a^2}=2b+7a\)

\(\Rightarrow b^2+5a^2=4b^2+49a^2+28ab\)

\(\Leftrightarrow\left(2a+b\right)\left(22a+3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2a=-b\\22a=-3b\end{cases}}\)

Với \(2a=-b\Rightarrow2\sqrt{x-1}=5-2x\)

\(\Rightarrow4\left(x-1\right)=25-20x+4x^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=3-\frac{\sqrt{7}}{2}\\x=3+\frac{\sqrt{7}}{2}\end{cases}}\)

Thử lại chỉ có \(x=3-\frac{\sqrt{7}}{2}\)thỏa mãn. 

Với \(22a=-3b\Rightarrow22\sqrt{x-1}=-3\left(2x-5\right)\)

\(\Rightarrow484\left(x-1\right)=9\left(2x-5\right)^2\)

\(\Leftrightarrow x=\frac{83}{9}\pm\frac{55\sqrt{7}}{18}\)

Thử lại đều không thỏa mãn.

ĐK: x \(\ge\)-4

Ta có: \(x^2+\sqrt{x+4}+\sqrt{x+1}=x+27\)

<=> \(x^2-x-27+\sqrt{x+4}+\sqrt{x+11}=0\)

<=> \(x^2-x-20+\sqrt{x+4}-3+\sqrt{x+11}-4=0\)

<=> \(\left(x-5\right)\left(x+4\right)+\frac{x+4-9}{\sqrt{x+4}+3}+\frac{x+11-16}{\sqrt{x+4}+4}=0\)

<=> \(\left(x-5\right)\left(x+4\right)+\frac{x-5}{\sqrt{x+4}+3}+\frac{x-5}{\sqrt{x+4}+4}=0\)

<=> \(\left(x-5\right)\left(x+4+\frac{1}{\sqrt{x+4}+3}+\frac{1}{\sqrt{x+11}+4}\right)=0\)

<=> \(x=5\)

(vì x \(\ge\)-4 => \(x+4\ge0\); \(\frac{1}{\sqrt{x+4}+3}>0\); \(\frac{1}{\sqrt{x+11}+4}>0\) 

=> \(x+4+\frac{1}{\sqrt{x+4}+3}+\frac{1}{\sqrt{x+11}+4}>0\))

Đk: \(-2\le x\le3\)

Ta có: \(\sqrt{x+2}-\sqrt{3-x}=x^2-6x+9\)

<=> \(\sqrt{x+2}-2-\sqrt{3-x}+1-x^2+6x-8=0\)

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

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

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

<=> \(\orbr{\begin{cases}x=2\\\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4=0\left(1\right)\end{cases}}\)

Vì \(-2\le x\le3\)

Do đó: \(\frac{1}{\sqrt{x+2}+2}>0\); \(\frac{1}{\sqrt{3-x}+1}>0\); 4 - x > 0

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

=> pt (1) vô nghiệm

Vậy S = {2}