Ta có : \(x+y\left(2+3x\right)=3\Leftrightarrow y=\frac{3-x}{3x+2}\)  ( vì x > 0 ) 

Khi đó : \(x+y=x+\frac{3-x}{3x+2}=\frac{3x^2+x+3}{3x+2}=A\) 

Chứng minh được :  \(A\ge\frac{-3+2\sqrt{11}}{3}\) => ... 

\(P\left(x\right)=3x^2-\left[3f\left(x\right)+1\right]x+3-f\left(x\right)=0\left(1\right)\)

Phương trình (1) có nghiệm thuộc \(\left(0;\frac{2}{3}\right)\) khi:

\(\hept{\begin{cases}\Delta=9f^2\left(x\right)+18f\left(x\right)-35\ge0\\P\left(0\right)=3-f\left(x\right)>0\\P\left(\frac{2}{3}\right)=\frac{11}{3}-3f\left(x\right)>0\end{cases}\Leftrightarrow\hept{\begin{cases}f\left(x\right)\le\frac{-3-2\sqrt{11}}{3}\left(h\right)f\left(x\right)\ge\frac{-3+2\sqrt{11}}{3}\\f\left(x\right)< 3\\f\left(x\right)< \frac{11}{9}\end{cases}}}\)

\(\Rightarrow f\left(x\right)\in(-\infty;\frac{-3-2\sqrt{11}}{3}]\)U\([\frac{-3+2\sqrt{11}}{3};\frac{11}{9})\)

Dễ thấy \(f\left(x\right)>0\forall x\in\left(0;\frac{2}{3}\right)\). Suy ra \(\frac{-3+2\sqrt{11}}{3}\le f\left(x\right)< \frac{11}{9}\)

Vậy \(minf\left(x\right)=\frac{-3+2\sqrt{11}}{3}\), đạt được khi \(x=\frac{-1+\sqrt{11}}{3}.\)

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

\(\ge3\sqrt[3]{\left[\frac{3}{4}\left(2x+1\right)\right]^2.\frac{2}{\left(2x+1\right)^2}}-\frac{3}{2}=\frac{3}{2}\sqrt[3]{9}-\frac{3}{2}\)

Dấu \(=\)khi \(\frac{3}{4}\left(2x+1\right)=\frac{2}{\left(2x+1\right)^2}\Leftrightarrow\left(2x+1\right)^3=\frac{8}{3}\Leftrightarrow x=\frac{1}{\sqrt[3]{3}}-\frac{1}{2}\).

\(f\left(x\right)=4x+\frac{3}{\left(x+1\right)^2}=2x+2+2x+2+\frac{3}{\left(x+1\right)^2}-4\ge3\sqrt[3]{\left(2x+2\right)^2.\frac{3}{\left(x+1\right)^2}}-4\)

\(=3\sqrt[3]{48}-4\)

Dấu \(=\)khi \(2x+2=\frac{3}{\left(x+1\right)^2}\Leftrightarrow\left(x+1\right)^3=\frac{3}{2}\Leftrightarrow x=\sqrt[3]{\frac{3}{2}}-1\).

\(f\left(x\right)=3x^2+\frac{8}{x}=3x^2+\frac{4}{x}+\frac{4}{x}\ge3\sqrt[3]{3x^2.\frac{4}{x}.\frac{4}{x}}=6\sqrt[3]{6}\)

Dấu \(=\)khi \(3x^2=\frac{4}{x}\Leftrightarrow x=\sqrt[3]{\frac{4}{3}}\).

1a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)

\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)

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

2. a) ĐK: x \(\ge\)4

Ta có: \(\sqrt{16x-64}=2\) <=> \(4\sqrt{x-4}=2\) <=> \(\sqrt{x-4}=\frac{1}{2}\)

<=> \(x-4=\frac{1}{4}\) <=> \(x=\frac{17}{4}\left(tm\right)\)

b) \(16x^2-\left(1+\sqrt{2}\right)^2=0\) <=> \(\left(4x-1-\sqrt{2}\right)\left(4x+1+\sqrt{2}\right)=0\)

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

c)Đk: x \(\ge\)0

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

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

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

<=> \(x=5+4\sqrt{3}\)

3a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{\sqrt{20}.\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}\)

\(=4\sqrt{5}-\frac{8\left(1+\sqrt{5}\right)}{5-1}=4\sqrt{5}-2\left(1+\sqrt{5}\right)=4\sqrt{5}-2-2\sqrt{5}=2\sqrt{5}-2\)

b) \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}+9\sqrt{2}}\)

\(=\frac{2\left(2\sqrt{2}-3\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{6}\left(\sqrt{5}+3\sqrt{3}\right)}=-\frac{2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=-\frac{3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)

$\sin18=\cos72=2 \cos^{2}36-1=2(1- \sin^{2}18)^{2}-1
\Leftrightarrow 8 \sin^{4}18 -8 \sin^{2}18- \sin18+1=0
\Leftrightarrow ( \sin18-1)[8 \sin^{3}18+8 \sin^{2}18-1]=0 $

ht