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

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

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

Để 2A > 3B hay   \(\frac{2\left(2+\sqrt{x}\right)}{\sqrt{x}}>3\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)

                          (2 +\(\sqrt{x}\)) (\(\frac{2}{\sqrt{x}}-\frac{3}{\sqrt{x}+1}\))   >0

                      vì \(\sqrt{x}+2>0\left(x>0\right)\)

\(\Rightarrow\frac{2}{\sqrt{x}}>\frac{3}{\sqrt{x}+1}\)

\(\Rightarrow\)\(2\sqrt{x}+2>3\sqrt{x}\)

  \(\sqrt{x}< 2\)

x <4

mà x>0 \(\Rightarrow\)0 <x<4

vậy để 2A >3b thì 0<x<4

#mã mã#

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^3+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)

\(=\frac{\sqrt{a^3}-3a\sqrt{b}+3\sqrt{a}.b-\sqrt{b^3}+2\sqrt{a^3}+\sqrt{b^3}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)

\(=\frac{3\sqrt{a^3}-3a\sqrt{b}+3b\sqrt{a}}{3\sqrt{a}\left(\sqrt{a^3}+\sqrt{b^3}\right)}+\frac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}\)

\(=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{\sqrt{a}+\sqrt{b}}=0\)

??? số đâu

Bạn xem lại đề bài.

Tam giác ABC vuông tại A. => AB<BC

Vì thế đề bài AB=BC là sai

\(A=\sqrt{\left(3+2\sqrt{3}\right)^2-5}=\sqrt{16+12\sqrt{3}}=2\sqrt{4+3\sqrt{3}}.\)

P/s: Đề có thể là như này số sẽ đẹp:

\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}=\sqrt{9-5-2\sqrt{3}}=\sqrt{4-2\sqrt{3}}\)\(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

\(B=\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}=2\sqrt{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}=2\)

\(\Leftrightarrow x^2+\left(2\sqrt{2}-3\right)x+4+3\sqrt{2}=0\)

\(\Delta=\left(2\sqrt{2}-3\right)^2-4\left(4+3\sqrt{2}\right)=1-24\sqrt{2}< 0\)

=> Pt vô nghiệm

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

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

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

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

TL:

= - 1

-HT-

!!!!!

trần đắc lợi lần sau nhớ gõ latex nha bạn, như này người làm dễ bị sai đề lắm

\(a\sqrt{b-1}+b\sqrt{a-1}\)

Áp dụng AM-GM :

\(a\sqrt{b-1}+b\sqrt{a-1}\)

\(=a\sqrt{1\cdot\left(b-1\right)}+b\sqrt{1\cdot\left(a-1\right)}\le a\cdot\frac{1+b-1}{2}+b\cdot\frac{1+a-1}{2}\)

\(=\frac{ab}{2}+\frac{ab}{2}=ab\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=2\)

cảm ơn nha