\(A=\frac{x-2+3\sqrt{x-2}+2}{x-2+4\sqrt{x-2}+3}\)  

Đặt: \(t=\sqrt{x-2}\ge0\)

\(A=\frac{\left(t+1\right)\left(t+2\right)}{\left(t+1\right)\left(t+3\right)}=\frac{t+2}{t+3}=1-\frac{1}{t+3}\ge1-\frac{1}{0+3}=\frac{2}{3}\)

Dấu "=" xảy ra <=> t = 0 hay x = 2

Vậy min A =2/3 tại x =2

Nè bạn :) 

Ta có : \(2ab+2ac\ge4a\sqrt{bc}\) (Cauchy_)

\(\Rightarrow a^2+2ab+2ac+4bc\ge a^2+4a\sqrt{bc}+4bc\)

\(\Rightarrow a^2+2ab+2ac+4bc\ge\left(a+2\sqrt{bc}\right)^2\)

\(\Rightarrow\sqrt{\left(a+2b\right)\left(a+2c\right)}\ge a+2\sqrt{bc}\)\(\left(1\right)\)

Tương tự : \(\sqrt{\left(b+2a\right)\left(b+2c\right)}\ge b+2\sqrt{ac}\)\(\left(2\right)\)

\(\sqrt{\left(c+2a\right)\left(c+2b\right)}\ge c+2\sqrt{ab}\)\(\left(3\right)\)

Từ \(\left(1\right);\left(2\right);\left(3\right)\)\(\Rightarrow\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge3\)

\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\ge\sqrt{3}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Thay vào biểu thức M ta được M = \(\frac{\sqrt{3}}{3}\)

\(\left(x-1\right)^2+3x=31\)

\(\Leftrightarrow x^2-2x+1+3x=31\)

\(\Leftrightarrow x^2+x-30=0\)

Ta có \(\Delta=1^2+4.30=121,\sqrt{\Delta}=11\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+11}{2}=5\\x=\frac{-1-11}{2}=-6\end{cases}}\)

\(\left(x-1\right)^2+3x=31\)

<=> x^2 -2x+1+3x=31

<=> x^2 +x+1=31

<=> x^2+x-30=0

<=> x^2 +6x-5x-30=0

<=> x(x+6)-5(x+6)=0

<=> (x+6)(x-5)=0

<=> x+6=0 hoặc x-5=0

<=> x=-6 hoặc x=5

Hint: Đặt \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\).