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

Toán Lớp 8: Hiệu hai bình phương A , 1/2x^2 – 3/2y^2 B , 3/4x^2 y^2 – 4y^4

Toán Lớp 8: Hiệu hai bình phương
A , 1/2x^2 – 3/2y^2
B , 3/4x^2 y^2 – 4y^4

Comments ( 2 )

  1. a)
    1/2x^2 – 3/2y^2
    = ( 1/(sqrt2)x – sqrt(3/2)y )( 1/(sqrt2)x + sqrt(3/2)y )
    Áp dụng : A^2-B^2=(A-B)(A+B)
    b)
    3/4x^2y^2 – 4y^4
    = y^2 ( 3/4x^2 – 4y^2 )
    = y^2 ( (sqrt3)/2x – 2y ) ( (sqrt3)/2x + 2y )
    Áp dụng : A^2-B^2=(A-B)(A+B)
     

  2. Hướng dẫn trả lời:
    a) 1/2x^2 – 3/2y^2
    = (1/sqrt2x)^2 – (sqrt{3/2}y)^2
    = (1/sqrt2x + sqrt{3/2}y)cdot(1/sqrt2x – sqrt{3/2}y)
    Giải thích:
    Áp dụng HĐT hiệu hai bình phương A^2 – B^2 = (A + B)cdot(A – B)
    b) 3/4x^2y^2 – 4y^4
    = y^2cdot(3/4x^2 – 4y^2)
    = y^2cdot[(sqrt3/2x)^2 – (2y)^2]
    = y^2cdot(sqrt3/2x + 2y)cdot(sqrt3/2x – 2y)
    Giải thích:
    Áp dụng HĐT A^2 – B^2 = (A + B)cdot(A – B)
    Giải đáp:
    a) = (1/sqrt2x + sqrt{3/2}y)cdot(1/sqrt2x – sqrt{3/2}y)
    b) = y^2cdot(sqrt3/2x + 2y)cdot(sqrt3/2x – 2y)

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