\(\frac{2}{4\sqrt{3}-7}+\frac{2}{4\sqrt{3}+7}\)

\(=-14-8\sqrt{3}+14-8\sqrt{3}\)

\(=-16\sqrt{3}\)

ta có \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3\)

ta có :\(\sqrt{4x+5}=\frac{\sqrt{9\left(4x+5\right)}}{3}\le\frac{9+4x+5}{2\times3}=\frac{2x+7}{3}\)

tương tự ta sẽ có  ; \(A\le\frac{2x+7}{3}+\frac{2y+7}{3}+\frac{2z+7}{3}=\frac{2}{3}\left(x+y+z\right)+7\le\frac{2}{3}\times3+7=9\)

Vậy GTLN của A=9

dấu bằng xảy ra khi x= y= z =1

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=3.3=9\)

\(\Rightarrow x+y+z\le3\).

\(A=\sqrt{4x+5}+\sqrt{4y+5}+\sqrt{4z+5}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(4x+5+4y+5+4z+5\right)}\)

\(=\sqrt{3\left[4\left(x+y+z\right)+15\right]}=9\)

Dấu \(=\)khi \(x=y=z=1\).

\(x^2+6x-3=4x\sqrt{2x-1}\left(1\right)\)      ĐK: \(x\ge\frac{1}{2}\)

Đặt \(\sqrt{2x-1}=a\ge0\)

\(\Rightarrow6x-3=3a^2\)

=> (1) <=> x^2 +3a^2 = 4ax

<=> x^2 -4ax +3a^2 =0

<=> x^2 -ax - 3ax +  3a^2 =0

<=> x(x-a) -3a(x-a) =0

<=> (x-a) ( x-3a ) =0

\(\Leftrightarrow\orbr{\begin{cases}x=a\\x=3a\end{cases}}\)

TH1: x=a

\(\Rightarrow x=\sqrt{2x-1}\)\(\left(x\ge0\right)\)

\(\Leftrightarrow x^2=2x-1\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

<=> x=1 (tm)

TH2: x= 3a

\(\Rightarrow x=3\sqrt{2x-1}\left(x\ge0\right)\)

\(\Leftrightarrow x^2=18x-9\)

\(\Leftrightarrow x^2-18x+9=0\)

\(\Delta=288\)

=> pt có 2 nghiệm pb \(\orbr{\begin{cases}x=\frac{18+12\sqrt{2}}{2}=9+6\sqrt{2}\left(tm\right)\\x=\frac{18-12\sqrt{2}}{2}=9-6\sqrt{2}\left(tm\right)\end{cases}}\)

Vậy ...

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