Question

[Nguyễn Thúy Hiền]

B1 \(A=\dfrac{9}{\sqrt{11}-\sqrt{2}}+\dfrac{\sqrt{22}-\sqrt{10}}{\sqrt{11}-\sqrt{5}}-\dfrac{22}{\sqrt{11}}\)

\(B=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}+\dfrac{a-b}{\sqrt{a}+\sqrt{b}}+\dfrac{2b}{\sqrt{b}}\left(a>b>0\right)\)

B2:Cho biểu thức:

\(P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1-\dfrac{1}{\sqrt{x}}\right)\)

a, Rút gọn P

b, Tính P khi x=\(\dfrac{1}{4}\)

c, Tìm tất cả các giá trị của x để P\(< 1\)

Answer 1

[Phương An]

Bài 1:

\(M=\dfrac{9}{\sqrt{11}-\sqrt{2}}-\dfrac{\sqrt{22}-\sqrt{10}}{\sqrt{11}-\sqrt{5}}-\dfrac{22}{\sqrt{11}}\)

\(=\dfrac{9\left(\sqrt{11}+\sqrt{2}\right)}{11-2}-\dfrac{\sqrt{2}\left(\sqrt{11}-\sqrt{5}\right)\left(\sqrt{11}+\sqrt{5}\right)}{11-5}-\dfrac{2.\left(\sqrt{11}\right)^2}{\sqrt{11}}\)

\(=\sqrt{11}+\sqrt{2}-\sqrt{2}-2\sqrt{11}=-\sqrt{11}\)

\(M=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}+\dfrac{a-b}{\sqrt{a}+\sqrt{b}}+\dfrac{2b}{\sqrt{b}}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}+\dfrac{2\left(\sqrt{b}\right)^2}{\sqrt{b}}\)

\(=\sqrt{a}-\sqrt{b}+\sqrt{a}-\sqrt{b}+2\sqrt{b}=2\sqrt{a}\)

Answer 2

[Phương An]

Bài 2:

a)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{\left(\sqrt{x}+1\right)+\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2}{\sqrt{x}+1}\) (*)

b)

Thay x = 0,25 vào (*), ta có:

\(M=\dfrac{2}{\sqrt{\dfrac{1}{4}}+1}=\dfrac{4}{3}\)

c)

\(M\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\ge1\)

\(\Leftrightarrow2\ge\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}\le1\)

\(\Leftrightarrow x\le1\)

mà x khác 1 và x > 0(theo ĐKXĐ)

=> 0 < x < 1 thì M \(\ge\) 1