a:
ĐKXĐ: x>=0
\(P=\dfrac{\sqrt{x}+3-4}{\sqrt{x}+3}=1-\dfrac{4}{\sqrt{x}+3}\)
\(\sqrt{x}+3>=3\)
=>\(\dfrac{4}{\sqrt{x}+3}< =\dfrac{4}{3}\)
=>\(-\dfrac{4}{\sqrt{x}+3}>=-\dfrac{4}{3}\)
=>\(P=\dfrac{-4}{\sqrt{x}+3}+1>=-\dfrac{4}{3}+1=-\dfrac{1}{3}\)
Dấu = xảy ra khi x=0
b: ĐKXĐ: x>=0
\(Q=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1+4}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)
=>\(Q>=2\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}-2=2\cdot2-2=2\)
Dấu = xảy ra khi \(\left(\sqrt{x}+1\right)^2=4\)
=>\(\sqrt{x}+1=2\)
=>x=1(nhận)
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