Bài 2 : 

a, \(\sqrt{0,09}=\sqrt{\frac{9}{100}}=\frac{3}{10}\)

b, \(\sqrt{-16}\)vô lí vì \(\sqrt{a}\)xảy ra khi a >= 0 mà -16 < 0 

c, \(\sqrt{0,25}.\sqrt{0,16}=\sqrt{\frac{25}{100}}.\sqrt{\frac{16}{100}}=\frac{5}{10}.\frac{4}{10}=\frac{20}{100}=\frac{1}{5}\)

e, \(\sqrt{\frac{4}{25}}=\frac{2}{5}\); f, \(\frac{6\sqrt{16}}{5\sqrt{0,04}}=\frac{24}{5\sqrt{\frac{4}{100}}}=\frac{24}{\frac{10}{10}}=24\)

g, \(\sqrt{0,36}-\sqrt{0,49}=\sqrt{\frac{36}{100}}-\sqrt{\frac{49}{100}}=\frac{6}{10}-\frac{7}{10}=-\frac{1}{10}\)

a, \(\sqrt{x-6}=13\)ĐK : x >= 6

\(\Leftrightarrow x-6=169\Leftrightarrow x=175\)

b, \(\sqrt{x^2-2x+4}=x-1\Leftrightarrow x^2-2x+4=x^2-2x+1\)

\(\Leftrightarrow4=1\)( vô lí ), vậy pt vô nghiệm 

c, \(\sqrt{x^2-8x+16}=9x-1\Leftrightarrow\left|x-4\right|=9x-1\)

ĐK : x >= 1/9

TH1 : \(x-4=9x-1\Leftrightarrow-8x=3\Leftrightarrow x=-\frac{3}{8}\)( ktm ) 

TH2 : \(x-4=1-9x\Leftrightarrow10x=5\Leftrightarrow x=\frac{1}{2}\)( tm )

c, \(\sqrt{x^2-x-4}=\sqrt{x-1}\Leftrightarrow x^2-x-4=x-1\)

\(\Leftrightarrow x^2-2x-3=0\Leftrightarrow x=3;x=-1\)

e, \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

TH1 : \(x-2=2x-3\Leftrightarrow x=1\)

TH2 : \(x-2=3-2x\Leftrightarrow3x=5\Leftrightarrow x=\frac{5}{3}\)

g, \(\sqrt{x+2\sqrt{x-1}}=2\)ĐK : x > = 1 

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

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)

TH1 : \(\sqrt{x-1}+1=2\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\)

TH2 : \(\sqrt{x-1}+1=-2\)( vô lí ) 

\(A=\frac{4\sqrt{x}}{\sqrt{x}+3}\left(\frac{x}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}+15}{x-9}\right)\)ĐK : \(x\ge0;x\ne9\)

\(=\frac{4\sqrt{x}}{\sqrt{x}+3}\left(\frac{x\sqrt{x}-3x+x+3\sqrt{x}+\sqrt{x}+15}{x-9}\right)\)

\(=\frac{4\sqrt{x}}{\sqrt{x}+3}\left(\frac{x\sqrt{x}-2x+4\sqrt{x}+15}{x-9}\right)\)

\(2a-2\sqrt{a}=2\sqrt{a}\left(\sqrt{a}-1\right)\)

Chúc học tốt 

ĐK \(x\le2\)Đặt \(\sqrt{2-x}=t\Rightarrow2-x=t^2\)\(\Rightarrow x=2-t^2\)ta có

\(N=2-t^2+t\)\(=-\left(t^2-2t\frac{1}{2}+\frac{1}{4}\right)+\frac{9}{4}=-\left(t-\frac{1}{2}\right)^2+\frac{9}{4}\ge\frac{9}{4}\)

Vì \(-\left(t-\frac{1}{2}\right)^2\le0\)

Dấu = xảy ra khi và chỉ khi \(t-\frac{1}{2}=0\Rightarrow t=\frac{1}{2}\Rightarrow\sqrt{2-x}=\frac{1}{2}\Leftrightarrow2-x=\frac{1}{4}\Rightarrow x=\frac{7}{4}\)tmđk

Vậy Max_N=9/4 <=> x=7/4

\(A=\sqrt{9x^2-12x+4}+1-3x=\sqrt{\left(3x-2\right)^2}+1-3x\)

\(=\left|3x-2\right|+1-3x\)Thay x = 1/3 vào A ta được : 

\(=\left|\frac{1}{3}.3-2\right|+1-\frac{3.1}{3}=1+1-1=1\)