Question

[Chún Hoàng]

giúp tớ bài 2 với ạ

Answer 1

[Nguyễn Huy Tú]

a, Ta có : \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)

Thay vào A ta được : \(A=\frac{\sqrt{2}-1-3}{\sqrt{2}-1+2}=\frac{\sqrt{2}-4}{\sqrt{2}+1}=\left(\sqrt{2}-4\right)\left(\sqrt{2}-1\right)=6-5\sqrt{2}\)

b, Với \(x\ge0;x\ne4;9\)

\(B=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{x-4}\)

\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\)

\(=\frac{3x-6\sqrt{x}}{x-4}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

c, Ta có : \(\frac{B}{A}< 1\Rightarrow\frac{\sqrt{x}-3}{3\sqrt{x}}< 1\Leftrightarrow\frac{\sqrt{x}-3}{3\sqrt{x}}-1< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-3-3\sqrt{x}}{3\sqrt{x}}< 0\Leftrightarrow\frac{-2\sqrt{x}-3}{3\sqrt{x}}< 0\)

\(\Rightarrow-2\sqrt{x}-3< 0\Leftrightarrow-2\sqrt{x}< 3\Leftrightarrow4x< 9\Leftrightarrow x< \frac{9}{4}\)

Kết hợp với đk vậy 0 =< x < 9/4 

d, \(\frac{\sqrt{x}-3}{3\sqrt{x}};3=\frac{9\sqrt{x}}{3\sqrt{x}}\Rightarrow9\sqrt{x}>\sqrt{x}-3\Rightarrow\frac{B}{A}< 3\)

Answer 2

[l҉o҉n҉g҉ d҉z҉]

a) Thay x vào A ta được : \(A=\frac{\sqrt{3-2\sqrt{2}}-3}{\sqrt{3-2\sqrt{2}+2}}=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}-3}{\sqrt{\left(\sqrt{2}-1\right)^2}+2}=\frac{\sqrt{2}-1-3}{\sqrt{2}-1+2}=\frac{-4+\sqrt{2}}{1+\sqrt{2}}\)

\(=\frac{\left(-4+\sqrt{2}\right)\left(1-\sqrt{2}\right)}{1-2}=\frac{-4+5\sqrt{2}-2}{-1}=6-5\sqrt{2}\)

b) \(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+\sqrt{5}x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+\sqrt{5}x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

c) \(\frac{B}{A}< 1\Leftrightarrow\frac{\frac{3\sqrt{x}}{\sqrt{x}+2}}{\frac{\sqrt{x}-3}{\sqrt{x}+2}}< 1\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}-3}< 1\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)

\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}< 0\Leftrightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)(1)

Vì \(2\sqrt{x}+3>0\)nên \(\left(1\right)\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)

Kết hợp với ĐK => Với \(\hept{\begin{cases}0\le x< 9\\x\ne4\end{cases}}\)thì B/A < 1

d) mình đang kẹt ý d) bạn thông cảm ;-;