\(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\)(ĐKXĐ: \(\frac{3}{2}\le x\le\frac{5}{2}\))

\(\Leftrightarrow2\sqrt{2x-3}+2\sqrt{5-2x}=6x^2-24x+28\)

\(\Leftrightarrow6x^2-24x+28-2\sqrt{2x-3}-2\sqrt{5-2x}=0\)

\(\Leftrightarrow\left(2x-3-2\sqrt{2x-3}+1\right)+\left(5-2x-2\sqrt{5-2x}+1\right)+6x^2-24x+24=0\)

\(\Leftrightarrow\left(\sqrt{2x-3}-1\right)^2+\left(\sqrt{5-2x}-1\right)^2+6\left(x-2\right)^2=0\)

Do \(\left(\sqrt{2x-3}-1\right)^2\ge0;\left(\sqrt{5-2x}-1\right)^2\ge0;6\left(x-2\right)^2\ge0\forall x\in R\)

Nên \(\hept{\begin{cases}\sqrt{2x-3}-1=0\\\sqrt{5-2x}-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-3=1\\5-2x=1\\x=2\end{cases}}\Leftrightarrow x=2\)(t/m ĐKXĐ)

Vậy pt có nghiệm duy nhất là x=2.

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

\(\Leftrightarrow9x^2-6x+5=\left(1-x^2\right)^2\)

\(\Leftrightarrow9x^2-6x+5=1-2x^2+x^4\)

\(\Leftrightarrow9x^2-6x+5-1+2x^2-x^4=0\)

\(\Leftrightarrow-x^4+11x^2-6x+4=0\)

\(\Leftrightarrow x^4-11x^2+6x-4=0\)

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

<=>\(\sqrt{\left(9x^2-6x+1\right)+4}=1-x^2\)

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

<=> 3x - 1 + 2 = 1 - x²

<=> 3x + x² = 1 +1 - 2

<=> x(3+x) = 0

<=> x = o hoặc 3+x =0 <=> x = -3

Vậy S= {0;-3}

Ta có:

\(P=\frac{1}{x+y}+\frac{xy+x+y}{x}+\frac{xy+x+y}{y}=\frac{1}{x+y}+1+\frac{y}{x}+y+1+\frac{x}{y}+x\)

\(=\frac{1}{x+y}+\left(x+y\right)+2+\left(\frac{x}{y}+\frac{y}{x}\right)\ge2+2+2=6\)

DẤU BẰNG XẢY RA:\(\Leftrightarrow x=y=\sqrt{2}-1\)