Bài 1:
1. Khi $x=25$ thì:
$A=\frac{\sqrt{25}+1}{25-3\sqrt{25}}=\frac{5+1}{25-3.5}=\frac{3}{5}$
2.
\(B=\frac{x}{\sqrt{x}(\sqrt{x}-3)}-\frac{\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-3)}=\frac{x-\sqrt{x}+2}{\sqrt{x}(\sqrt{x}-3)}\)
\(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-3)}: \frac{x-\sqrt{x}+2}{\sqrt{x}(\sqrt{x}-3)}=\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-3)}.\frac{\sqrt{x}(\sqrt{x}-3)}{x-\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+1}{x-\sqrt{x}+2}\)
3.
Áp dụng BĐT Cô-si:
$x+1\geq 2\sqrt{x}$
$\Rightarrow x-\sqrt{x}+2=(x+1)-\sqrt{x}+1\geq 2\sqrt{x}-\sqrt{x}+1=\sqrt{x}+1$
$\Rightarrow P=\frac{\sqrt{x}+1}{x-\sqrt{x}+2}\leq \frac{\sqrt{x}+1}{\sqrt{x}+1}=1$
Vậy $P_{\max}=1$ khi $x=1$
Bài 2.
a.
Khi $x=\sqrt{2}$ thì $A=\frac{2}{\sqrt{2}-1}=\frac{2(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=2(\sqrt{2}+1)$
b.
\(B=\frac{x+2\sqrt{x}}{x+\sqrt{x}-2}-1=\frac{x+2\sqrt{x}-(x+\sqrt{x}-2)}{x+\sqrt{x}-2}\)
\(=\frac{\sqrt{x}+2}{x+\sqrt{x}-2}=\frac{\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{1}{\sqrt{x}-1}\)
c.
\(P=A:B=\frac{2}{x-1}: \frac{1}{\sqrt{x}-1}=\frac{2(\sqrt{x}-1)}{x-1}=\frac{2(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2}{\sqrt{x}+1}\)
Vì $\sqrt{x}\geq 0$ với mọi $x$ thuộc đkxđ
$\Rightarrow \sqrt{x}+1\geq 1$
$\Rightarrow P=\frac{2}{\sqrt{x}+1}\leq 2$
Vậy $P_{\max}=2$. Giá trị này đạt tại $x=0$
Bài III:
a. Khi $x=\frac{9}{16}$ thì $\sqrt{x}=\frac{3}{4}$
Khi đó:
$B=\frac{\frac{3}{4}-3}{2}=\frac{-9}{8}$
2.
$A=\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\frac{\sqrt{x}+11}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{2\sqrt{x}+14}{(\sqrt{x}-3)(\sqrt{x}+3)}$
$M=AB=\frac{2(\sqrt{x}+7)}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{2}=\frac{\sqrt{x}+7}{\sqrt{x}+3}$
3.
$M=1+\frac{4}{\sqrt{x}+3}$
Vì $\sqrt{x}\geq 0$ với mọi $x$ thuộc đkxđ nên $\sqrt{x}+3\geq 3$
$\Rightarrow \frac{4}{\sqrt{x}+3}\leq \frac{4}{3}$
$\Rightarrow M=1+\frac{4}{\sqrt{x}+3}\leq 1+\frac{4}{3}=\frac{7}{3}$
Vậy $M_{\max}=\frac{7}{3}$ khi $x=0$
Bài 4 bạn cũng làm tương tự thôi. Mấu chốt ở chỗ rút gọn đúng.
