Question

Cho P = \(\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)RútgonP.Cho\sqrt{a}+\sqrt{b}=4.Tim.GTNNcuaP\)

Answer 1

sorry, câu b nhầm \(\sqrt{a}+\sqrt{b}=4\) thành \(a+b=4\)

Sửa:

Có \(\sqrt{a}+\sqrt{b}=4\Rightarrow a+b+2\sqrt{ab}=16\Leftrightarrow a+b=16-2\sqrt{ab}\)

Áp dụng BĐT cô si cho 2 số ko âm

\(a+b\ge2\sqrt{ab}\)\(\Rightarrow16-2\sqrt{ab}\ge2\sqrt{ab}\Leftrightarrow16\ge4\sqrt{ab}\)

\(\Leftrightarrow-\sqrt{ab}\ge-4\)

"="\(\Leftrightarrow a=b=4\)

Answer 2

a/ ĐKXĐ: a,b\(\ge\) 0, ab\(\ne\) 1

\(P=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)+\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)-ab+1}{ab-1}\right]:\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)-\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)+ab-1}{ab-1}\right]\)

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

\(P=\frac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}.\frac{ab-1}{-2\sqrt{a}-2}=\frac{2\sqrt{ab}\left(\sqrt{a+1}\right)}{-2\left(\sqrt{a}+1\right)}=-\sqrt{ab}\)

b/ BĐT cô si cho 2 số ko âm

\(a+b\ge2\sqrt{ab}\Rightarrow-\left(a+b\right)\le-2\sqrt{ab}\)

\(\Leftrightarrow-4\le-2\sqrt{ab}\Leftrightarrow-\sqrt{ab}\ge-2\)

"="\(\Leftrightarrow a=b=2\)