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

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

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

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

??? 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

\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\)\(\Rightarrow\frac{|4x+3|}{|x+1|}=3\)

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

TH1 : \(4x+3=-3\left(x+1\right)\)

\(\Rightarrow4x+3=-3x-3\)

\(\Rightarrow7x=-6\Leftrightarrow x=-\frac{6}{7}\)

Th2 : \(4x+3=3\left(x+1\right)\)

\(\Rightarrow4x+3=3x+3\)

\(\Rightarrow x=0\)

Vậy \(x\in\left\{0;-\frac{6}{7}\right\}\)

\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\Rightarrow\frac{|4x+3|}{|x+1|}=9\)

\(\Rightarrow|4x+3|=9|x+1|\)

\(TH1:|4x+3|=-9|x+1|\)

\(\Rightarrow4x+3=-9x-9\Rightarrow13x=-12\Rightarrow x=-\frac{12}{13}\)

\(TH2:|4x+3|=9|x+1|\)

\(\Rightarrow4x+3=9x+9\)

\(\Rightarrow5x=-6\Rightarrow x=\frac{-6}{5}\)