minh@@22@22222@@@@? 

Ta có : 

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

Để P đạt GTNN thì \(1-\frac{2}{\sqrt{x}+1}\) phải đạt GTNN hay \(\frac{2}{\sqrt{x}+1}>0\) và đạt GTLN \(\Rightarrow\)\(\sqrt{x}+1>0\) và đạt GTNN 

\(\Rightarrow\)\(\sqrt{x}+1=1\)

\(\Rightarrow\)\(\sqrt{x}=0\)

\(\Rightarrow\)\(x=0\)

Suy ra : 

\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{0}-1}{\sqrt{0}+1}=\frac{-1}{1}=-1\)

Vậy \(P_{min}=-1\) khi \(x=0\)

\(3^{x+1}.9^{x+2}.27^x=81^x\)

\(3^x.3.3^{2\left(x+2\right)}.3^{3x}=3^{4x}\)

\(3^{x+1+2x+4+3x}=3^{4x}\)

\(\Rightarrow x+1+2x+4+3x=4x\)

\(6x+5-4x=0\)

\(2x+5=0\)

\(2x=-5\Rightarrow x=-\dfrac{5}{2}\)

a)

\(A=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)

\(A-2=-\dfrac{3}{x^2-8x+22}=-\dfrac{3}{\left(x-4\right)^2+6}\ge-\dfrac{3}{6}=-\dfrac{1}{2}\)

\(A\ge\dfrac{3}{2}\) khi x =4

\(a=\left(x+1\right)\left(x+2\right)\left(x+8\right)\left(x+9\right)\)

\(a=\left[\left(x+1\right)\left(x+9\right)\right]\left[\left(x+2\right)\left(x+8\right)\right]\)

\(a=\left[x\left(x+9\right)+1\left(x+9\right)\right]\left[x\left(x+8\right)+2\left(x+8\right)\right]\)

\(a=\left(x^2+9x+x+9\right)\left(x^2+8x+2x+16\right)\)

\(a=\left(x^2+10x+9\right)\left(x^2+10x+16\right)\)

\(a=\left(x^2+10x+12,5-3,5\right)\left(x^2+10x+12,5+3,5\right)\)

\(a=\left(x^2+10x+12,5\right)^2-\dfrac{49}{4}\ge-\dfrac{49}{4}\)