\(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\left(dkxd:x>0,x\ne9\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-\left(2x-\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
Ta có : \(P=A+\dfrac{1}{B}=\dfrac{x+7}{\sqrt{x}}+\left(1:\dfrac{\sqrt{x}}{\sqrt{x}-3}\right)=\dfrac{x+7}{\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{x+7+\sqrt{x}-3}{\sqrt{x}}=\dfrac{x+\sqrt{x}+4}{\sqrt{x}}\) \(=1+\left(\sqrt{x}+\dfrac{4}{\sqrt{x}}\right)\left(x>0\right)\)
Áp dụng BĐT Cosi, ta có :
\(\sqrt{x}+\dfrac{4}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{4}{\sqrt{x}}}=2\sqrt{4}=4\)
Dấu '' = '' xảy ra khi \(\sqrt{x}=\dfrac{4}{\sqrt{x}}\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy \(min_P=4\) khi và chỉ khi \(x=4\)
mọi người giúp mình giải câu 2 3 4 và phần hình học vs ạ. mình cảm ơnnnnnnnnnnnnn
