Question

2\sqrt2+\sqrt{50} -\sqrt{98}=???как решить?

Answer 1

$2\sqrt2+\sqrt{50} -\sqrt{98} \\ 2 \sqrt{2} + 5 \sqrt{2} - 7 \sqrt{2} \\ 7\sqrt{2} - 7 \sqrt{2} \\ 0 \times \sqrt{2 } = 0$
ОТВЕТ:0

Answer 2

$2\sqrt{2} +\sqrt{50}-\sqrt{98} =2\sqrt{2}+\sqrt{25*2}-\sqrt{49*2}=2\sqrt{2}+5\sqrt{2}-7\sqrt{2} =7\sqrt{2}-7\sqrt{2}=0$