$\frac{√26}{√13}$ - $\frac{2+√2 }{√2+ 1}$ - $\frac{7 }{√2 - 3}$

2 câu trả lời

$\displaystyle \begin{array}{{>{\displaystyle}l}} \frac{\sqrt{26}}{\sqrt{13}} -\frac{2+\sqrt{2}}{\sqrt{2} +1} -\frac{7}{\sqrt{2} -3}\\ =\sqrt{\frac{26}{13}} -\frac{\sqrt{2}\left(\sqrt{2} +1\right)}{\sqrt{2} +1} +\frac{7\left( 3+\sqrt{2}\right)}{3-2}\\ =\sqrt{2} -\sqrt{2} +21+7\sqrt{2}\\ =21+7\sqrt{2} \ \\ \end{array}$

Lời giải:

 \(\begin{array}{l}\dfrac{\sqrt{26}}{\sqrt{13}}-\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{7}{\sqrt{2}-3}\\=\sqrt{\dfrac{26}{13}}-\dfrac{\sqrt{2}(\sqrt{2}+1)}{\sqrt{2}+1}+\dfrac{7(3+\sqrt{2})}{9-2}\\=\sqrt{2}-\sqrt{2}+(3+\sqrt{2})\\=3+\sqrt{2}\end{array}\)

Câu hỏi trong lớp Xem thêm