Đặt \(\hept{\begin{cases}\sqrt[3]{x+2}=a\\\sqrt[3]{5-x}=b\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a+b=1\\a^3+b^3=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\\left(a+b\right)\left(a^2-ab+b^2\right)=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\a^2-ab+b^2=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\\left(a+b\right)^2-3ab=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\1-3ab=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\ab=-2\end{cases}}\)

Làm nốt

Ta có: \(\sqrt[3]{x+2}+\sqrt[3]{5-x}=1\)

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

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

<=> \(3\sqrt[3]{\left(x+2\right)\left(5-x\right)}=-6\)

<=> \(\sqrt[3]{\left(x+2\right)\left(5-x\right)}=-2\)

<=>\(3x+10-x^2=-8\)

<=> \(x^2-3x-18=0\)

<=> \(x^2-6x+3x-18=0\)

<=> \(\left(x-6\right)\left(x+3\right)=0\)

<=> \(\orbr{\begin{cases}x=6\\x=-3\end{cases}}\)

Đk: \(-3\le x\le\frac{13}{4}\)

Ta có: \(2\sqrt{x+3}+\sqrt{13-4x}=x^2+4x+2\)

<=> \(2\left(\sqrt{x+3}-2\right)+\left(\sqrt{13-4x}-3\right)-x^2-4x+5=0\)

<=> \(2\cdot\frac{x+3-4}{\sqrt{x+3}+2}+\frac{13-4x-9}{\sqrt{13-4x}+3}-\left(x-1\right)\left(x+5\right)=0\)

<=> \(2\cdot\frac{x-1}{\sqrt{x+3}+2}-\frac{4x-4}{\sqrt{13-4x}+3}-\left(x-1\right)\left(x+5\right)=0\)

<=> \(\left(x-1\right)\left(\frac{2}{\sqrt{x+3}+2}-\frac{4}{\sqrt{13-4x}+3}-x-5\right)=0\)

<=> \(\orbr{\begin{cases}x=1\\\frac{2}{\sqrt{x+3}+2}-\frac{4}{\sqrt{13-4x}+3}-x-5=0\left(1\right)\end{cases}}\)

Do \(-3\le x\le\frac{13}{4}\)

=> \(\frac{2}{\sqrt{x+3}+2}\le1\); \(-\frac{4}{\sqrt{13-4x}+3}< 0\); \(-x-5< -\left(-3\right)-5=-2\)

=> \(\frac{2}{\sqrt{x+3}+2}-\frac{4}{\sqrt{13-4x}+3}-x-5< 1-2=-1< 0\)

=>  pt (1) vô nghiệm

Vậy S = {1}

Cách 1:

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

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

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

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

\(\Leftrightarrow\sqrt{2x-1}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Cách 2:

Điều kiện: \(x\ge\frac{1}{2}\)

Ta có:
\(VT=\sqrt{4x-1}+\sqrt{4x^2-1}\ge\sqrt{4.\frac{1}{2}-1}+\sqrt{4.\left(\frac{1}{2}\right)^2-1}=1=VP\)

Dấu = xảy ra khi \(x=\frac{1}{2}\)

 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 \(S=\left\{\frac{2}{3};\frac{5}{3}\right\}\)