Ta có P=\(\frac{20-x-5\sqrt{x}+4\sqrt{x}}{\sqrt{x}+5}\)

P=\(\frac{\sqrt{x}\left(4-\sqrt{x}\right)+5\left(4-\sqrt{x}\right)}{\sqrt{x}+5}\)

P=\(\frac{\left(\sqrt{x}+5\right).\left(4-\sqrt{x}\right)}{\sqrt{x}+5}\)

P=\(4-\sqrt{x}\)

b) Ta có P=\(4-\sqrt{x}\)\(\le\)4 với mọi x\(\ge0\)

=> P đạt GTLN là 4 khi \(\sqrt{x}=0\)

                                      => x=0

\(A=4\left(x+2\right)-\left(2x+1\right)\left(2x-1\right)\)

\(A=4x+8-4x^2+1\)

\(A=-\left(4x^2-4x+4\right)+13\)

\(A=-\left(2x+2\right)^2+13\)

Vì \(-\left(2x+2\right)^2\le0\forall x\)

\(\Rightarrow-\left(2x+2\right)^2+13\le13\forall x\)

\(\Rightarrow A_{max}=13\)khi và chỉ khi \(-\left(2x+2\right)^2=0\Rightarrow x=-1\)

ĐKXĐ: x>=0

\(A=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x-3}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\dfrac{x-3}{\sqrt{x}+1}\)

\(=\dfrac{x-\sqrt{x}-x+3}{\sqrt{x}+1}=\dfrac{-\sqrt{x}+3}{\sqrt{x}+1}\)

\(=\dfrac{-\sqrt{x}-1+4}{\sqrt{x}+1}=-1+\dfrac{4}{\sqrt{x}+1}\)

\(\dfrac{4}{\sqrt{x}+1}< =\dfrac{4}{1}=4\)

=>\(\dfrac{4}{\sqrt{x}+1}-1< =4-1=3\)

Dấu = xảy ra khi x=0

a, \(M=\frac{3\left(x^2+1\right)}{\left(x^4+x^2\right)+\left(2x^3+2x\right)+\left(6x^2+6x\right)}=\frac{3\left(x^2+1\right)}{x^2\left(x^2+1\right)+2x\left(x^2+1\right)+6\left(x^2+1\right)}=\frac{3\left(x^2+1\right)}{\left(x^2+2x+6\right)\left(x^2+1\right)}=\frac{3}{x^2+2x+6}\)

b, ta có: \(M=\frac{3}{x^2+2x+6}=\frac{3}{\left(x^2+2x+1\right)+5}=\frac{3}{\left(x+1\right)^2+5}\)

Vì \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+5\ge5\Rightarrow\frac{1}{\left(x+1\right)^2+5}\le\frac{1}{5}\Rightarrow M=\frac{3}{\left(x+1\right)^2+5}\le\frac{3}{5}\)

Dấu "=" xảy ra <=>x+1=0 <=> x=-1

`A=(1/(x-sqrtx)+1/(sqrtx-1)):(sqrtx+1)/(sqrtx-1)^2`

`=((sqrtx+1)/(x-sqrtx)).(sqrtx-1)^2/(sqrtx+1)`

`=(sqrtx-1)^2/(x-sqrtx)`

`=(sqrtx-1)/sqrtx`