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

Toán Lớp 8: Được câu nào làm câu đấy ạ! Cũng cần nhanh chút ạ! Cho x^2 = y^2 + z^2. Chøng minh rằng: a) ( 5x – 3y + 4z) ( 5x – 3y – 4z) = (3x -5y)^

Toán Lớp 8: Được câu nào làm câu đấy ạ! Cũng cần nhanh chút ạ!
Cho x^2 = y^2 + z^2. Chøng minh rằng:
a) ( 5x – 3y + 4z) ( 5x – 3y – 4z) = (3x -5y)^2
b) ( 13x – 12 y – 5z)( 13x – 12y + 5z) = ( 12x – 13y)^2

Comments ( 1 )

  1. Giải đáp:
    Ta có: x^2 = y^2 + z^2
    ⇒ x^2 – y^2 – z^2 = 0
    a, (5x – 3y + 4z)(5x – 3y – 4z) = (3x – 5y)^2
    ⇔ (5x – 3y)^2 – (4z)^2 = (3x – 5y)^2
    ⇔ (5x – 3y)^2 – (4z)^2 – (3x – 5y)^2 = 0
    ⇔ (5x)^2 – 2.5x.3y + (3y)^2 – 16z^2 – [(3x)^2 – 2.3x.5y + (5y)^2] = 0
    ⇔ 25x^2 – 30xy + 9y^2 – 16z^2 – (9x^2 – 30xy + 25y^2) = 0
    ⇔ 25x^2 – 30xy + 9y^2 – 16z^2 – 9x^2 + 30xy – 25y^2 = 0
    ⇔ 16x^2 – 16z^2 – 16y^2 = 0
    ⇔ 16(x^2 – z^2 – y^2) = 0
    ⇔ 16 . 0 = 0
    Vậy (5x – 3y + 4z)(5x – 3y – 4z) = (3x – 5y)^2
    b, (13x – 12y – 5z)(13x – 12y + 5z) = (12x – 13y)^2
    ⇔ (13x – 12y)^2 – (5z)^2 = (12x – 13y)^2
    ⇔ (13x – 12y)^2 – (5z)^2 – (12x – 13y)^2 = 0
    ⇔ (13x)^2 – 2.13x.12y + (12y)^2 – (5z)^2 – [(12x)^2 – 2.12x.13y + (13y)^2] = 0
    ⇔ 169x^2 – 312xy + 144x^2 – 25z^2 – (144x^2 – 312xy + 169y^2) = 0
    ⇔ 169x^2 – 312xy + 144x^2 – 25z^2 – 144x^2 + 312xy – 169y^2 = 0
    ⇔ 25x^2 – 25z^2 – 25y^2 = 0
    ⇔ 25(x^2 – z^2 – y^2) = 0
    ⇔ 25.0 = 0
    → đpcm
    Vậy (13x – 12y – 5z)(13x – 12y + 5z) = (12x – 13y)^2

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