rút gọn a ² + √a trên a – √a +1 , trừ 2a+ √a trên √a cộng 1 15/08/2021 Bởi Autumn rút gọn a ² + √a trên a – √a +1 , trừ 2a+ √a trên √a cộng 1
$\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1$ $(ĐKXĐ:a \ne 0)$ $=\dfrac{\sqrt[]{a}(a \sqrt[]{a}+1)}{a- \sqrt[]{a} +1}-\dfrac{ \sqrt[]{a}+(2 \sqrt[]{a}+1}{ \sqrt[]{a}}+1$ $= \sqrt[]{a}( \sqrt[]{a} +1)-(2 \sqrt[]{a}+1)+1$ $=a- \sqrt[]{a}= \sqrt[]{a}( \sqrt[]{a}-1)$ Vậy $\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1=\sqrt[]{a}( \sqrt[]{a}-1)$ Bình luận
$\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1$ $(ĐKXĐ:a \ne 0)$
$=\dfrac{\sqrt[]{a}(a \sqrt[]{a}+1)}{a- \sqrt[]{a} +1}-\dfrac{ \sqrt[]{a}+(2 \sqrt[]{a}+1}{ \sqrt[]{a}}+1$
$= \sqrt[]{a}( \sqrt[]{a} +1)-(2 \sqrt[]{a}+1)+1$
$=a- \sqrt[]{a}= \sqrt[]{a}( \sqrt[]{a}-1)$
Vậy $\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1=\sqrt[]{a}( \sqrt[]{a}-1)$