a/ \(A=\frac{1}{5+2\sqrt{6-x^2}}\)

Có: \(-x^2\le0\)với mọi x

=> \(6-x^2\le6\)

=> \(0\le\sqrt{6-x^2}\le\sqrt{6}\)

=> \(5\le5+2\sqrt{6-x^2}\le5+2\sqrt{6}\)

=> \(\frac{1}{5+2\sqrt{6}}\le\frac{1}{5+2\sqrt{6-x^2}}\le\frac{1}{5}\); với mọi x

=> \(\hept{\begin{cases}maxA=\frac{1}{5}\Leftrightarrow\sqrt{6-x^2}=0\Leftrightarrow x=\pm\sqrt{6}\\minA=\frac{1}{5+2\sqrt{6}}\Leftrightarrow\sqrt{6-x^2}=\sqrt{6}\Leftrightarrow x=0\end{cases}}\)

Vậy:...

b/ \(B=\sqrt{-x^2+2x+4}=\sqrt{-\left(x-1\right)^2+5}\)

Có: \(-\left(x-1\right)^2\le0\)với mọi x

=> \(-\left(x-1\right)^2+5\le5\)

=> \(0\le\sqrt{-\left(x-1\right)^2+5}\le\sqrt{5}\)

=> \(0\le B\le\sqrt{5}\)với mọi x

=> \(\hept{\begin{cases}maxB=\sqrt{5}\Leftrightarrow-\left(x-1\right)^2=0\Leftrightarrow x=1\\minB=0\Leftrightarrow\left(x-1\right)^2=5\Leftrightarrow x=\pm\sqrt{5}+1\end{cases}}\)

Vậy:...

a)Ta có:

\(0\le2\sqrt{6-x^2}\le2\sqrt{6}\)

\(\Leftrightarrow\frac{1}{5}\ge\frac{1}{5+2\sqrt{6-x^2}}\ge\frac{1}{5+2\sqrt{6}}=5-2\sqrt{6}\)

\(\Rightarrow\hept{\begin{cases}MAX\left(A\right)=\frac{1}{5}\\MIN\left(A\right)=5-2\sqrt{6}\end{cases}}\)Dấu "=" xảy ra khi \(\hept{\begin{cases}x=0\left(MIN\right)\\x=\sqrt{6}\left(MAX\right)\end{cases}}\)

b)Ta có:

\(0\le B=\sqrt{-x^2+2x+4}=\sqrt{5-\left(x-1\right)^2}\le\sqrt{5}\)

\(\Rightarrow\hept{\begin{cases}MAX\left(B\right)=\sqrt{5}\\MIN\left(B\right)=0\end{cases}}\)Dấu "=" xảy ra khi \(x=1\left(MAX\right)\) và \(\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}\left(MIN\right)}\)

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2

a) đk: \(x\ge0;x\ne1\)

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

\(A=\frac{x+2+\left(\sqrt{x}-1\right)\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\div\frac{\sqrt{x}-1}{2}\)

\(A=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{2}{\sqrt{x}-1}\)

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

\(A=\frac{2}{x+\sqrt{x}+1}\)

c) Ta có: \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\frac{1}{4}\right)+\frac{3}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\) 

=> \(\frac{2}{x+\sqrt{x}+1}>0\left(\forall x\ne1\right)\)

d) Ta chỉ có thể tìm GTLN thôi

Để A đạt GTLN => \(x+\sqrt{x}+1\) phải đạt GTNN

Dấu "=" xảy ra khi: \(x=0\)

Vậy Max(A) = 2 khi x = 0