Question

A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) - \(\dfrac{3\sqrt{x}+1}{x-1}\)

a) Rút gọn A

b) Tính giá trị của A khi x = 4 - \(2\sqrt{3}\)

c) Tìm x để A = \(\dfrac{1}{2}\)

d) Tìm x để A < 1

e) Tìm x ∈ Z để A nhận giá trị nguyên

f) Tìm GTNN của A

Answer 1

a, ĐK: \(x\ge0,x\ne1\)

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{x-1}\)

\(=\dfrac{x+1+2\sqrt{x}+x+1-2\sqrt{x}-3\sqrt{x}-1}{x-1}\)

\(=\dfrac{2x-3\sqrt{x}+1}{x-1}\)

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

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

Answer 2

b, \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

Khi đó: 

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

\(=\dfrac{2\left(\sqrt{3}-1\right)-1}{\left(\sqrt{3}-1\right)+1}\)

\(=\dfrac{2\sqrt{3}-3}{\sqrt{3}}\)

\(=2-\sqrt{3}\)

Answer 3

c, \(A=\dfrac{1}{2}\)

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

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}+1\)

\(\Leftrightarrow3\sqrt{x}=3\)

\(\Leftrightarrow x=1\left(l\right)\)

Vậy không tồn tại giá trị x thỏa mãn \(A=\dfrac{1}{2}\).

Answer 4

d, \(A< 1\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\)

\(\Leftrightarrow2\sqrt{x}-1< \sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow\left\{{}\begin{matrix}0\le x< 4\\x\ne1\end{matrix}\right.\)

Answer 5

e, \(A\in Z\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\in Z\)

\(\Leftrightarrow2-\dfrac{3}{\sqrt{x}+1}\in Z\)

\(\Leftrightarrow\sqrt{x}+1\inƯ_3=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow x\in\left\{0;4\right\}\)