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

Toán Lớp 9: 1, 3 + 2√2 2, 3 – 2√2 3, 6 + 2√5 4, 6 – 2√5 5, 7 + 2√6 6, 7 – 2√6 7, 5 + 2√6 8, 5 – 2√6 9, 6 + 4√2 10, 6 – 4√2

Home/ Toán Lớp 9: 1, 3 + 2√2 2, 3 – 2√2 3, 6 + 2√5 4, 6 – 2√5 5, 7 + 2√6 6, 7 – 2√6 7, 5 + 2√6 8, 5 – 2√6 9, 6 + 4√2 10, 6 – 4√2

Toán Lớp 9: 1, 3 + 2√2 2, 3 – 2√2 3, 6 + 2√5 4, 6 – 2√5 5, 7 + 2√6 6, 7 – 2√6 7, 5 + 2√6 8, 5 – 2√6 9, 6 + 4√2 10, 6 – 4√2

Tháng Mười Hai 14, 2022 Toán học 1 Comment 0 views

Toán Lớp 9: 1, 3 + 2√2
2, 3 – 2√2
3, 6 + 2√5
4, 6 – 2√5
5, 7 + 2√6
6, 7 – 2√6
7, 5 + 2√6
8, 5 – 2√6
9, 6 + 4√2
10, 6 – 4√2

Comments ( 1 )

  1. thaolanphuobg
    thaolanphuobg
    14 Tháng Mười Hai, 2022 at 8:54 sáng
    Reply
    Giải đáp:
     
    Lời giải và giải thích chi tiết:
    1/  3+2\sqrt{2}
    =2+1+2\sqrt{2}
    =(\sqrt{2})^2+2\sqrt{2}+(1)^2
    =(\sqrt{2}+1)^2
    2/  3-2\sqrt{2}
    =2+1-2\sqrt{2}
    =(\sqrt{2})^2-2\sqrt{2}+(1)^2
    =(\sqrt{2}-1)^2
    3/  6+2\sqrt{5}
    =5+1+2\sqrt{2}
    =(\sqrt{5})^2+2\sqrt{5}+(1)^2
    =(\sqrt{5}+1)^2
    4/  6-2\sqrt{5}
    =5+1+2\sqrt{2}
    =(\sqrt{5})^2-2\sqrt{5}+(1)^2
    =(\sqrt{5}-1)^2
    5/  7+2\sqrt{6}
    =6+1+2\sqrt{2}
    =(\sqrt{6})^2+2\sqrt{6}+(1)^2
    =(\sqrt{6}+1)^2
    6/  7-2\sqrt{6}
    =6+1-2\sqrt{2}
    =(\sqrt{6})^2-2\sqrt{6}+(1)^2
    =(\sqrt{6}-1)^2
    7/  5+2\sqrt{6}
    =\frac{2(5+2\sqrt{6})}{2}
    =\frac{10+4\sqrt{6}}{2}
    =\frac{6+4+4\sqrt{6}}{2}
    =\frac{(\sqrt{6})^2+2.2\sqrt{6}+(2)^2}{2}
    =\frac{(\sqrt{6}+2)^2}{2}
    8/  5-2\sqrt{6}
    =\frac{2(5-2\sqrt{6})}{2}
    =\frac{10-4\sqrt{6}}{2}
    =\frac{6+4-4\sqrt{6}}{2}
    =\frac{(\sqrt{6})^2-2.2\sqrt{6}+(2)^2}{2}
    =\frac{(\sqrt{6}-2)^2}{2}
    9/  6+4\sqrt{2}
    =2+4+2\sqrt{2}
    =(\sqrt{2})^2+2.2\sqrt{2}+(2)^2
    =(\sqrt{2}+2)^2
    10/  6-4\sqrt{2}
    =2+4-2\sqrt{2}
    =(\sqrt{2})^2-2.2\sqrt{2}+(2)^2
    =(\sqrt{2}-2)^2

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

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