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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Bài 1: Rút gọn b) √$\frac{3}{2}$ – $\frac{2 √2}{ √3}$ – $\frac{1}{4}$.√$6$ c) $\frac{2}{ √2+ √3}$ + $\frac{3}{ √3- √2}$ d)

Toán Lớp 9: Bài 1: Rút gọn
b) √$\frac{3}{2}$ – $\frac{2 √2}{ √3}$ – $\frac{1}{4}$.√$6$
c) $\frac{2}{ √2+ √3}$ + $\frac{3}{ √3- √2}$
d) √$7$-$4 √3$ + √$7+4 √3$ ( dấu √ của cả $4 √3$)

Comments ( 1 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    b) \sqrt{3/2} – (2\sqrt{2})/\sqrt{3} – 1/4 \sqrt{6}
    = 1/2 \sqrt{6} – 2/3 \sqrt{6} – 1/4 \sqrt{6}
    = (1/2-2/3-1/4) \sqrt{6}
    = -5/12 \sqrt{6}
    c) 2/(\sqrt{2}+\sqrt{3}) + 3/(\sqrt{3}-\sqrt{2})
    = (2(\sqrt{2}-\sqrt{3}))/((\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})) + (3(\sqrt{3}+\sqrt{2}))/((\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}))
    = (2(\sqrt{2}-\sqrt{3}))/(2-3) + (3(\sqrt{3}+\sqrt{2}))/(3-2)
    = (-2(\sqrt{2}-\sqrt{3}))/1 + (3(\sqrt{3}+\sqrt{2}))/1
    = -2 (\sqrt{2}-\sqrt{3}) + 3 (\sqrt{3}+\sqrt{2})
    = -2\sqrt{2} + 2\sqrt{3} + 3\sqrt{3} + 3\sqrt{2}
    = \sqrt{2} + 5 \sqrt{3}
    d) \sqrt{7-4\sqrt{3}} + \sqrt{7+4\sqrt{3}}
    = \sqrt{(2-\sqrt{3})^2} + \sqrt{(2+\sqrt{3})^2}
    = |2-\sqrt{3}| + |2+\sqrt{3}|
    = 2 – \sqrt{3} + 2 + \sqrt{3}
    = 4
     

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222-9+11+12:2*14+14 = ? ( )