Question

Nêu điều kiện xác định và rút gọn biểu thức:B=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

Answer 1

đk : x >= 0 ; x khác 1 

\(B=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}=\dfrac{4\sqrt{x}}{x-1}\)

Answer 2

B xác định \(< =>\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}=\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{x-1}=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}=\dfrac{4\sqrt{x}}{x-1}\)

Answer 3

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\left(luôn.đúng\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

Answer 4

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

ĐKXĐ: x>0; x<>1

Answer 5

\(ĐK:x>1\)

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

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

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

\(=\dfrac{4\sqrt{x}}{x-1}\)

Answer 6

\(Đk:x\ge0,x\ne1\)

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

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

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

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

\(B=\dfrac{4\sqrt{x}}{x-1}\)