\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}.\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}=\dfrac{x\left(\sqrt{x}-1\right)+16.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{\left(\sqrt{x}-1\right).\left(x+16\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
ĐKXĐ : \(x\ge0\) ; \(x\ne1\) ; \(x\ne9\)
b) ta có : \(A=\dfrac{x+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)
Theo BĐT Cosi có:
\(\left(\sqrt{x}+3\right)+\dfrac{25}{\sqrt{x}+3}\ge2.\sqrt{\left(\sqrt{x}+3\right).\dfrac{25}{\sqrt{x}+3}}\)
=> \(\left(\sqrt{x}+3\right)+\dfrac{25}{\sqrt{x}+3}\ge2\sqrt{25}\)
=> \(\left(\sqrt{x}+3\right)+\dfrac{25}{\sqrt{x}+3}-6\ge10-6\)
=> \(A\ge4\)
=> MinA = 4
Dấu = xảy ra <=> \(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\)
<=> \(\left(\sqrt{x}+3\right)^2=25\)
<=> \(\sqrt{x}+3=5\)
<=> \(\sqrt{x}=2\)
<=> x = 4
