b) $\dfrac{a.\sqrt{a} + b.\sqrt{b}}{\sqrt{a} + \sqrt{b}} - \sqrt{ab} = (\sqrt{a} - \sqrt{b})^2$
Edited by Phạm Hữu Bảo Chung, 22-09-2011 - 19:01.
Edited by Phạm Hữu Bảo Chung, 22-09-2011 - 19:01.
Edited by Phạm Quang Toàn, 27-09-2011 - 17:24.
Edited by Phạm Quang Toàn, 27-09-2011 - 17:27.
Xem cho kĩ ông ơi$\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=\dfrac{(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=a-2\sqrt{ab}+b=(\sqrt{a}+\sqrt{b})^2$
Edited by Phạm Hữu Bảo Chung, 22-09-2011 - 19:06.
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