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.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: ((√x-2)/(x-1) -(√x+2)/(x+2√x+1) ).((1-x)^2)/2

Toán Lớp 9: ((√x-2)/(x-1) -(√x+2)/(x+2√x+1) ).((1-x)^2)/2

Comments ( 1 )

  1. Điều kiện xác định $x\ge 0, x\ne 1$
    $\begin{array}{l} \left( {\dfrac{{\sqrt x  – 2}}{{x – 1}} – \dfrac{{\sqrt x  + 2}}{{x + 2\sqrt x  + 1}}} \right).\dfrac{{{{\left( {1 – x} \right)}^2}}}{2}\\  = \left[ {\dfrac{{\sqrt x  – 2}}{{\left( {\sqrt x  – 1} \right)\left( {\sqrt x  + 1} \right)}} – \dfrac{{\sqrt x  + 2}}{{{{\left( {\sqrt x  + 1} \right)}^2}}}} \right].\dfrac{{{{\left( {1 – x} \right)}^2}}}{2}\\  = \dfrac{{\left( {\sqrt x  – 2} \right)\left( {\sqrt x  + 1} \right) – \left( {\sqrt x  + 2} \right)\left( {\sqrt x  – 1} \right)}}{{{{\left( {\sqrt x  + 1} \right)}^2}\left( {\sqrt x  – 1} \right)}}.\dfrac{{{{\left( {x – 1} \right)}^2}}}{2}\\  = \dfrac{{x – \sqrt x  – 2 – \left( {x + \sqrt x  – 2} \right)}}{{\left( {x – 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{{{\left( {x – 1} \right)}^2}}}{2}\\  = \dfrac{{ – 2\sqrt x }}{{\left( {x – 1} \right)\left( {\sqrt x  + 1} \right)}}.\dfrac{{{{\left( {x – 1} \right)}^2}}}{2}\\  = \dfrac{{ – \sqrt x \left( {x – 1} \right)}}{{\sqrt x  + 1}} = \dfrac{{ – \sqrt x \left( {\sqrt x  – 1} \right)\left( {\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}\\  =  – \sqrt x \left( {\sqrt x  – 1} \right) = \sqrt x  – x \end{array}$  

Leave a reply

222-9+11+12:2*14+14 = ? ( )