Toán Lớp 9: A = ($\frac{2\sqrt{x} + x}{x\sqrt{x} -1}$ – $\frac{1}{\sqrt{x} – 1}$) : ($\frac{x – 1}{x + \sqrt{x} + 1}$)
Rút gọn biểu thức A
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 9: A = ($\frac{2\sqrt{x} + x}{x\sqrt{x} -1}$ – $\frac{1}{\sqrt{x} – 1}$) : ($\frac{x – 1}{x + \sqrt{x} + 1}$)
Rút gọn biểu thức A
Comments ( 1 )
Dkxd:x \ge 0;x \ne 1\\
A = \left( {\dfrac{{2\sqrt x + x}}{{x\sqrt x – 1}} – \dfrac{1}{{\sqrt x – 1}}} \right):\dfrac{{x – 1}}{{x + \sqrt x + 1}}\\
= \left[ {\dfrac{{2\sqrt x + x}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}} – \dfrac{1}{{\sqrt x – 1}}} \right].\dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2\sqrt x + x – \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ – \sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ – 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{1}{{1 – x}}
\end{array}$