1/ ĐKXĐ: \(x\ge 0\\x=4(TM)\\→A=\dfrac{\sqrt 4+1}{\sqrt 4+2}=\dfrac{2+1}{2+2}=\dfrac{3}{4}\)
2/ ĐKXĐ: \(x\ge 0; x\ne 1\)
\(B=\dfrac{3}{\sqrt x-1}-\dfrac{\sqrt x+5}{x-1}\\=\dfrac{3(\sqrt x+1)}{(\sqrt x-1)(\sqrt x+1)}-\dfrac{\sqrt x+5}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{3\sqrt x+3-\sqrt x-5}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{2\sqrt x-2}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{2(\sqrt x-1)}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{2}{\sqrt x+1}\)
→ĐPCM
3/ \(2.A.B=2.\dfrac{\sqrt x+1}{\sqrt x+2}.\dfrac{2}{\sqrt x+1}=\dfrac{4}{\sqrt x+2}\\→P=\dfrac{4}{\sqrt x+2}+\sqrt x\\=\dfrac{4}{\sqrt x+2}+\sqrt x+2-2\)
Áp dụng bất đẳng thức Cô-si với 2 số dương \(\dfrac{4}{\sqrt x+2}\) và \(\sqrt x+2\)
\(→\dfrac{4}{\sqrt x+2}+\sqrt x+2\ge 2\sqrt{\dfrac{4}{\sqrt x+2}.(\sqrt x+2)}=2.\sqrt 4=4\\→P≥4-2=2\\→\min P=2\)
\(→\) Dấu "=" xảy ra khi \(\dfrac{4}{\sqrt x+2}=\sqrt x+2\\↔(\sqrt x+2)^2=4\\↔\sqrt x+2=2(vì\,\,\sqrt x+2>0∀x)\\↔\sqrt x=0\\↔x=0(TM)\)
Vậy \(\min P=2\) khi \(x=0\)
