rút gọn bt ($\frac{x+2}{x√x+1}$ – $\frac{1}{√x + 1}$ ). $\frac{4√x}{3}$ với x$\geq$ 0 01/07/2021 Bởi Everleigh rút gọn bt ($\frac{x+2}{x√x+1}$ – $\frac{1}{√x + 1}$ ). $\frac{4√x}{3}$ với x$\geq$ 0
$\begin{array}{l} \left( {\dfrac{{x + 2}}{{x\sqrt x + 1}} – \dfrac{1}{{\sqrt x + 1}}} \right).\dfrac{{4\sqrt x }}{3}\\ = \left[ {\dfrac{{x + 2}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}} – \dfrac{1}{{\sqrt x + 1}}} \right].\dfrac{{4\sqrt x }}{3}\\ = \dfrac{{x + 2 – \left( {x – \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\ = \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\ = \dfrac{{4\sqrt x }}{{3\left( {x – \sqrt x + 1} \right)}} \end{array}$ Bình luận
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))` Bình luận
$\begin{array}{l} \left( {\dfrac{{x + 2}}{{x\sqrt x + 1}} – \dfrac{1}{{\sqrt x + 1}}} \right).\dfrac{{4\sqrt x }}{3}\\ = \left[ {\dfrac{{x + 2}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}} – \dfrac{1}{{\sqrt x + 1}}} \right].\dfrac{{4\sqrt x }}{3}\\ = \dfrac{{x + 2 – \left( {x – \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\ = \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\ = \dfrac{{4\sqrt x }}{{3\left( {x – \sqrt x + 1} \right)}} \end{array}$
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))`