A=?;B=?

a)

\(M=2+\sqrt{\left(2x\right)^2-2.2x.3+3^2}\)

\(\Rightarrow M=2+\sqrt{\left(2x-3\right)^2}\)

\(\Rightarrow M=2+2x-3\)

\(\Rightarrow M=2x-1\)

b)

(+) x=5/2

=> \(M=2.\frac{5}{2}-1=5-1=4\)

(+) x= - 1/5

=> \(M=2.\frac{\left(-1\right)}{5}-1=-\frac{2}{5}-1=-\frac{7}{5}\)

đk: x>=0; x khác 3

a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

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

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

ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)

a.

\(B=\left(\frac{x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\left(\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\\ =\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

b. Ta có :

\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\\ =\sqrt{25+2\cdot5\cdot\sqrt{2}+2}-\sqrt{16+2\cdot4\cdot\sqrt{2}+2}\\ =\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\\ =5+\sqrt{2}-4-\sqrt{2}=1\)

\(B=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{1+1}{1+3}=\frac{2}{4}=\frac{1}{2}\)

c. Giả sử B>\(\frac{1}{3}\), ta có

\(B=\frac{\sqrt{x}+1}{\sqrt{x}+3}>\frac{1}{3}\\ \Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}+3}-\frac{1}{3}>0\\ \Leftrightarrow\\\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow\frac{2\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\left(luondungvoix>0\right)\)

Vậy.........

\(=\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right):\left(\frac{\left(\sqrt{x}-4\right)\left(x+1\right)-\sqrt{x}\left(1-x\right)}{1-x^2}\right)\)

\(=\left(\frac{x+2-x-\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{x\sqrt{x}+\sqrt{x}-4x-4-\sqrt{x}+x\sqrt{x}}{1-x^2}\right)\)

\(=\frac{2-\sqrt{x}}{\sqrt{x}+1}:\frac{2x\sqrt{x}-3x-4}{\left(1-x\right)\left(1+x\right)}\)

\(=\frac{2-\sqrt{x}}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(1+x\right)}{2x\sqrt{x}-3x-4}\)

\(=\frac{\left(2-\sqrt{x}\right)\left(\sqrt{x}+x\sqrt{x}-1+x\right)}{2x\sqrt{x}-3x-4}\)

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

tới đêy tự xử đi

a) \(B=\sqrt{\frac{\left(x-2\right)^4}{\left(3-x\right)^2}}+\frac{x^2+1}{x-3}=\frac{\left(x-2\right)^2}{3-x}+\frac{x^2+1}{x-3}\\ =\frac{-\left(x-2\right)^2+x^2+1}{x-3}=\frac{-x^2+4x-4+x^2+1}{x-3}=\frac{4x-3}{x-3}\)

b) khi x=0,5 thì

\(B=\frac{4\cdot0,5-3}{0,5-3}=\frac{2}{5}\)