rút gọn bt ($\frac{x+2}{x√x+1}$ – $\frac{1}{√x + 1}$ ). $\frac{4√x}{3}$ với x$\geq$ 0

0 bình luận về “rút gọn bt ($\frac{x+2}{x√x+1}$ – $\frac{1}{√x + 1}$ ). $\frac{4√x}{3}$ với x$\geq$ 0”

1. $\begin{array}{l} \left( {\dfrac{{x + 2}}{{x\sqrt x  + 1}} – \dfrac{1}{{\sqrt x  + 1}}} \right).\dfrac{{4\sqrt x }}{3}\\  = \left[ {\dfrac{{x + 2}}{{\left( {\sqrt x  + 1} \right)\left( {x – \sqrt x  + 1} \right)}} – \dfrac{1}{{\sqrt x  + 1}}} \right].\dfrac{{4\sqrt x }}{3}\\  = \dfrac{{x + 2 – \left( {x – \sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {x – \sqrt x  + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\  = \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {x – \sqrt x  + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\  = \dfrac{{4\sqrt x }}{{3\left( {x – \sqrt x  + 1} \right)}} \end{array}$  

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2. Đáp án:

   `((x+2)/(xsqrtx+1)-1/(sqrtx+1)).(4sqrtx)/3(x>=0)`

   `=((x+2)/(xsqrtx+1)-(x-sqrtx+1)/(xsqrtx+1)).(4sqrtx)/3`

   `=(x+2-x+sqrtx-1)/(xsqrtx+1).(4sqrtx)/3`

   `=(sqrtx+1)/(xsqrtx+1).(4sqrtx)/3`

   `=1/(x-sqrtx+1).(4sqrtx)/3`

   `=(4sqrtx)/(3(x-sqrtx+1))`

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