b)gọi BT trên là P
\(P=\frac{x+16}{\sqrt{x}+3}=\frac{x-9+25}{\sqrt{x+3}}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)
Vì\(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3>0\Rightarrow\frac{25}{\sqrt{x}+3}>0\)
Áp dụng BĐT Cô-si cho 2 số không âm \(\sqrt{x}+3\) và \(\frac{25}{\sqrt{x}+3}\) ta có:
\(\sqrt{x}+3+\frac{15}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right)\frac{25}{\sqrt{x}+3}}=10\\ \sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\ge4\\ \Rightarrow P\ge4\)
Dấu "=' xảy ra khi \(\left(\sqrt{x}+3\right)^2=25\Rightarrow x=4\)
Vậy \(P_{min}=4\) khi \(x=4\)
gọi BT ở trên là P
a)\(P=\frac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\frac{2\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\sqrt{x}+3}\\ P=\frac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\\ P=\frac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)} \\ P=\frac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\\ P=\frac{\left(\sqrt{x}-1\right)\left(x+16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x+16}{\sqrt{x}+3}\)
