Mình bổ sung lại đề bài nha mọi người bị thiếu
Tìm điều kiện xác định rồi rút gọn nhé:
a) $A=\frac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\frac{2\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\sqrt{x}+3}\\=\frac{x\sqrt{x}+26\sqrt{x}-19}{(\sqrt{x}+1)^2-4}-\frac{2\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\sqrt{x}+3}\\=\frac{x\sqrt{x}+26\sqrt{x}-19}{(\sqrt{x}+1-2)(\sqrt{x}+1+2)}-\frac{2\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\sqrt{x}+3}$
$=\frac{x\sqrt{x}+26\sqrt{x}-19}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{2\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\sqrt{x}+3}\\=\frac{x\sqrt{x}+26\sqrt{x}-19}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}+\frac{(\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}\\=\frac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-\sqrt{x}-3\sqrt{x}+3}{(\sqrt{x}-1)(\sqrt{x}+3)}\\=\frac{x\sqrt{x}+16\sqrt{x}-x-16}{(\sqrt{x}-1)(\sqrt{x}+3)}$
$=\frac{\sqrt{x}(x+16)-(x+16)}{(\sqrt{x}-1)(\sqrt{x}+3)}\\=\frac{(\sqrt{x}-1)(x+16)}{(\sqrt{x}-1)(\sqrt{x}+3)}\\=\frac{x+16}{\sqrt{x}+3}$
b) Ta có: $\frac{x+16}{\sqrt{x}+3}\\=\frac{x-9+25}{\sqrt{x}+3}\\=\frac{(\sqrt{x}-3)(\sqrt{x}+3)+25}{\sqrt{x}+3}\\=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\\=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6$
Áp dụng bdt cô-si cho 2 số không âm ta có:
$\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\geq 2\sqrt{(\sqrt{x}+3).\frac{25}{\sqrt{x}+3}}\\\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\geq 2\sqrt{25}\\\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\geq 10\\\Rightarrow P=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\geq 10-6\\\Rightarrow P\geq 4$
Vậy ..........
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