Tìm GTNN của bt \(\frac{x+3}{\sqrt{x}}\)
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\(Q=\sqrt{x}+\dfrac{25}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{25}{\sqrt{x}}}=10\)
Dấu = xảy ra khi x=25
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{\left(x+7\right)}{\sqrt{x}+3}=\)\(\frac{\left(x+3\sqrt{x}\right)-3\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}\)\(=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}\)
\(=\left(\sqrt{x}+3\right)+\frac{16}{\sqrt{x}+3}-6\)\(\ge8-6=2\)(AM-GM)
''='' <=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
P\(=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-3\)
ap dung cosi cho 2 so duong \(\left(\sqrt{x}+3\right)va\frac{16}{\sqrt{x}+3}taduoc\)
\(\sqrt{x}+3+\frac{16}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right)\frac{16}{\sqrt{x}+3}}\)\(\ge2\sqrt{16}=8\)
\(\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-3\ge8-3=5\)
dau = xay ra <=> \(\left(\sqrt{x}+3\right)=\frac{16}{\sqrt{x}+3}\)
<=> x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x>0;x\ne9\)
\(A=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)