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

Toán Lớp 8: Chứng minh đẳng thức: ${a}^{5} – {b}^{5} – {(a – b)}^{5} = 5ab(a – b)({a}^{2} – ab + {b}^{2})$

Toán Lớp 8: Chứng minh đẳng thức:
${a}^{5} – {b}^{5} – {(a – b)}^{5} = 5ab(a – b)({a}^{2} – ab + {b}^{2})$

Comments ( 2 )

  1. Giải đáp:
    $\\$
    Xét vế trái :
    a^5 – b^5 – (a-b)^5
    = a^5 – b^5 – (a-b)^3 (a-b)^2
    = a^5 – b^5 – (a^3 – 3a^2b + 3ab^2 – b^3) (a^2 – 2ab + b^2)
    = a^5 – b^5 – (a^2 – 2ab +b^2 ) (a^3 – 3a^2b + 3ab^2 – b^3)
    = a^5 – b^5 – (a^5 – 5a^4b + 10a^3b^2 -10a^2b^3 + 5ab^4 -b^5)
    = a^5 – b^5 – a^5 + 5a^4b – 10a^3b^2 + 10a^2b^3 – 5ab^4 + b^5
    = (a^5 – a^5) + (-b^5 – b^5) + 5a^4b – 10a^3b^2 + 10a^2b^3 – 5ab^4
    = 5a^4b – 10a^3b^2 + 10a^2b^3 – 5ab^4
    = 5ab (a^3 – 2a^2b + 2ab^2 – b^3) (1)
    $\\$
    Xét vế phải :
    5ab (a-b) (a^2 – ab +b)
    = 5ab [a (a^2-ab+b^2) -b(a^2-ab+b)]
    = 5ab [a^3 – a^2b+ ab^2 – a^2b + ab^2 – b^3]
    = 5ab [a^3 + (-a^2b – a^2b) + (ab^2 + ab^2)-b^3]
    = 5ab [a^3 – 2a^2b + 2ab^2 – b^3] (2)
    $\\$
    Từ (1), (2)
    -> a^5 – b^5 – (a-b)^5 = 5ab (a-b) (a^2-ab+b^2)
     

  2. ${a}^{5} – {b}^{5} – {(a – b)}^{5} = 5ab(a – b)({a}^{2} – ab + {b}^{2})$
     Chứng minh vế trái: 
    $\left(a-b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)$
    $=a^5-b^5-\left(a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5\right)$
    $=a^5-b^5-a^5+5a^4b-10a^3b^2+10a^2b^3-5ab^4+b^5$
    $=5a^4b-10a^3b^2+10a^2b^3-5ab^4$
    $=5ab\left(a^3-2a^2b+2ab^2-b^3\right)$
    Thật vậy, vế trái bằng: 5ab[(a-b)(a^2-ab+b^2)]=5ab(a^3-2a^2b+2ab^2-b^3)

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

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