Ta có : \(\frac{23\sqrt{2}}{\sqrt{2}+\sqrt{14+5\sqrt{3}}}=\frac{46}{2+\sqrt{28+10\sqrt{3}}}=\frac{46}{2+\sqrt{\left(\sqrt{3}+5\right)^2}}=\frac{46}{7+\sqrt{3}}\)

\(=\frac{46\left(7-\sqrt{3}\right)}{\left(7+\sqrt{3}\right)\left(7-\sqrt{3}\right)}=\frac{46\left(7-\sqrt{3}\right)}{46}=7-\sqrt{3}\)

Suy ra a = 7 , b = -1

=> a x b = -7

Giải:

Đề=

=\(\sqrt{\left(\sqrt{3}-3\right)^2.\frac{\left(-1\right)}{\sqrt{3}-3}}\)

\(=\sqrt{3-\sqrt{3}}\)

1) Thay x=16 vào A ta có:

A=\(\frac{16+\sqrt{16}+1}{\sqrt{16}+2}\)

A=\(\frac{16+4+1}{4+2}\)

A=\(\frac{21}{6}=\frac{7}{2}\)

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

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

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

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

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

\(3,P=A.B=\frac{x+\sqrt{x}+1}{\sqrt{x}+2}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Ta thấy \(\left(\sqrt{x}-1\right)^2>0\Rightarrow x-2\sqrt{x}+1>0\)

\(\Rightarrow x+\sqrt{x}+1>3\sqrt{x}\)

\(\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>\frac{3\sqrt{x}}{\sqrt{x}}\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>3\left(đpcm\right)\)

a: Khi x=16 thì \(A=\dfrac{4+1}{4-1}=\dfrac{5}{3}\)

b: \(P=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{x-4}=\dfrac{x+\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\)

Để P lớn nhất thì căn x-2=1

=>căn x=3

=>x=9

a) Từ đề bài suy ra A = \(\frac{3\sqrt{x}-6}{x-2\sqrt{x}}+\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}\) ĐKXĐ của A, B là: x > 0 ; x ≠ 4

A = \(\frac{3\sqrt{x}-6+\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

A = \(\frac{3\sqrt{x}-6+x-5\sqrt{x}+6+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

b) Ta có P = A.B = \(\frac{\sqrt{x}-1}{\sqrt{x}-2}.\frac{\sqrt{x}-2}{\sqrt{x}+9}\) = \(\frac{\sqrt{x}-1}{\sqrt{x}+9}\) Suy ra ĐK: x ≥ 1 (x ∈ Z)

\(\sqrt{\frac{\sqrt{x}-1}{\sqrt{x}+9}}< \frac{1}{3}\) ⇔ \(\frac{\sqrt{x}-1}{\sqrt{x}+9}< \frac{1}{9}\) ⇔ \(9\sqrt[]{x}-9< \sqrt{x}+9\) ⇔ \(8\sqrt{x}< 18\) ⇔1 ≤ \(x< \frac{81}{16}\)

a. Ta có: \(A=\sqrt{\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2006^2}\right)}=\sqrt{\frac{1}{2}.\frac{3}{2}.\frac{2}{3}.\frac{4}{3}...\frac{2015}{2016}.\frac{2017}{2016}}\)

\(=\sqrt{\frac{1}{2}.\frac{2017}{2016}}=\sqrt{\frac{2017}{4032}}\) 

b. Với b > 0 thì a > 0, ta có: \(B=\frac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{b}-\frac{\sqrt{a}}{\sqrt{b}}=\frac{-\sqrt{b}}{\sqrt{b}}=-1\)

Với b < 0 thì a < 0, ta có: \(B=\frac{\sqrt{ab}-\sqrt{b^2}}{b}-\frac{\sqrt{ab}}{\sqrt{b^2}}=\frac{\sqrt{ab}-\sqrt{b^2}}{b}+\frac{\sqrt{ab}}{b}=\frac{2\sqrt{ab}+b}{b}\)