Toán Lớp 9: Tính :
5$\sqrt{2a}$ – 2$\sqrt{18a}$ +3$\sqrt{72a}$ +$\sqrt{a}$ với a$\geq$ 0
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Toán Lớp 9: Tính :
5$\sqrt{2a}$ – 2$\sqrt{18a}$ +3$\sqrt{72a}$ +$\sqrt{a}$ với a$\geq$ 0
Comments ( 2 )
5\sqrt{2a}-2\sqrt{18a}+3\sqrt{72a}+\sqrt{a}(a>=0)
=5\sqrt{2a}-2.3\sqrt{2a}+3.6\sqrt{2a}+\sqrt{a}
=5\sqrt{2a}-6\sqrt{2a}+18\sqrt{2a}+\sqrt{a}
=17\sqrt{2a}+\sqrt{a}
5\sqrt(2a)-2\sqrt(18a)+3\sqrt(72a)+\sqrta(a>=0)
=5\sqrt(2a)-2\sqrt(3^2 . 2a)+3\sqrt(6^2 . 2a)+\sqrta
=5\sqrt(2a)-6\sqrt( 2a)+18\sqrt( 2a)+\sqrta
=17\sqrt(2a)+\sqrta