\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=\sqrt{5}\left(2-5-4+11\right)=4\sqrt{5}\)

\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-\sqrt{5.25}-\sqrt{5.16}+\sqrt{5.121}\)

\(=2\sqrt{5}-\sqrt{5}.\sqrt{25}-\sqrt{5}.\sqrt{16}+\sqrt{5}.\sqrt{121}\)

\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=4\sqrt{5}\)

Ta có x + y + z = 2

=> (x + y + z)² = 4

<=> x² + y² + z² + 2(xy + yz + zx) = 4

<=> 2 + 2(xy + yz + zx) = 4

<=> xy + yz + zx = 1

<=> \(\frac{xyz}{x}+\frac{xyz}{y}+\frac{xyz}{z}=1\)

<=>  \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\left(\text{đpcm}\right)\)

ĐK \(-2\ge x\le2\)

Ta có \(9x^2+2\sqrt{x^2-4}=36\)

\(\Leftrightarrow9\left(x^2-4\right)+2\sqrt{x^2-4}=0\)

Đặt \(\sqrt{x^2-4}=t\left(t\ge0\right)\Rightarrow x^2-4=t^2\)ta có 

\(9t^2+2t=0\Leftrightarrow t\left(9t+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=0\\9t+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=0\left(TM\text{Đ}K\right)\\t=-\frac{2}{9}\left(lo\text{ại}\right)\end{cases}}}\)

\(\Leftrightarrow\sqrt{x^2-4}=0\Leftrightarrow x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\text{Đ}K\right)\)

điều kiện \(x\ge0;P\ge0\)

Để chứng minh \(p>\sqrt{P}\)luôn đúng ta cần chứng minh P>1 luôn đúng.

Giả sử P>1 \(\Leftrightarrow\)\(\frac{x+16}{\sqrt{x}+3}>1\)\(\Leftrightarrow\)\(x+16>\sqrt{x}+3\)\(\Leftrightarrow\)\(x-\sqrt{x}+13>0\)

\(\Leftrightarrow\)\(x+\sqrt{x}+\frac{1}{4}+12,75>0\)\(\Leftrightarrow\)\(\left(\sqrt{x}+\frac{1}{2}\right)^2+12,75>0\)luôn luôn đúng

như vậy P luôn luôn >1 là đúng\(\Leftrightarrow\)\(p>\sqrt{P}\)luôn đúng (đpcm)

Vì \(p\)là số nguyên tổ nên tổng các ước nguyên dương của \(p^4\)là \(1+p+p^2+p^3+p^4\).

Đặt \(p^4+p^3+p^2+p+1=n^2\)

\(\Leftrightarrow4p^4+4p^3+4p^2+4p+1=4n^2\)

Ta có: 

\(4p^4+4p^3+4p^2+4p+4>4p^4+4p^3+p^2=\left(2p^2+p\right)^2\)

\(4p^4+4p^3+4p^2+4p+4< 4p^4+4p^3+9p^2+4p+4=\left(2p^2+p+2\right)^2\)

Suy ra \(\left(2p^2+p\right)^2< 4n^2< \left(2p^2+p+2\right)^2\)

\(\Rightarrow\left(2n\right)^2=\left(2p^2+p+1\right)^2=4p^4+4p^3+5p^2+2p+1\)

\(\Rightarrow p^2-2p-3=0\)

\(\Leftrightarrow\left(p+1\right)\left(p-3\right)=0\)

\(\Rightarrow p=3\)thỏa mãn. 

Vậy \(p=3\).

Ta có: \(4ab\le2a^2+2b^2\)

=> \(\sqrt{2a^2+7b^2+16ab}\le\sqrt{4a^2+9b^2+12ab}=\sqrt{\left(2a+3b\right)^2}=2a+3b\)

=> \(\frac{a^2}{\sqrt{2a^2+7b^2+16ab}}\ge\frac{a^2}{2a+3b}\)

Chứng minh tương tự 

=> \(T\ge\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\)

Áp dụng bđt bunhia dạng phân thức

=> \(T\ge\frac{\left(a+b+c\right)^2}{2a+3b+2b+3c+2c+3a}=\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=1\)

=> \(MinT=1\)xảy ra khi a=b=c=5/3