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

Toán Lớp 8: (a+b)(b+c)(c+a)+4abc=c(a+b)^2+a(b+c)^2+b(a+c)^2

Toán Lớp 8: (a+b)(b+c)(c+a)+4abc=c(a+b)^2+a(b+c)^2+b(a+c)^2

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết !
    to Chứng minh:
    (a+b)(b+c)(c+a)+4abc = c(a+b)^2+a(b+c)^2+b(a+c)^2
    Ta có:
    @ c(a+b)^2+a(b+c)^2+b(a+c)^2
    = c(a^2+2ab+b^2)+a(b^2+2bc+c^2)+b(a^2+2ac+c^2)
    = a^2c+2abc+b^2c+ab^2+2abc+ac^2+a^2b+2abc+bc^2
    = a^2c+ac^2+b^2c+bc^2+ab^2+a^2b+(2abc+2abc+2abc)
    = a^2c+ac^2+b^2c+bc^2+ab^2+a^2b+6abc (1)
    @  (a+b)(b+c)(c+a)+4abc
    = (ab+ac+b^2+bc)(c+a)+4abc
    = abc+a^2b+ac^2+a^2c+b^2c+ab^2+bc^2+abc+4abc
    = a^2c+ac^2+b^2c+bc^2+ab^2+a^2b+(abc+abc+4abc)
    = a^2c+ac^2+b^2c+bc^2+ab^2+a^2b+6abc (2)
    Từ (1); (2) ta có:
    => (a+b)(b+c)(c+a)+4abc = c(a+b)^2+a(b+c)^2+b(a+c)^2
     

  2. $\text{YeunhatbanT}$ 
    (a + b)(b + c)(c + a)  +4abc
    =(ab + ac + b^2 + bc)(c + a) + 4abc
    =abc  + a^2b + ac^2 + a^2c + b^2c + b^2a + bc^2 + abc + 4abc
    =a^2b + ac^2 + a^2c + b^2c + b^2a + bc^2 + 6abc
    c(a+b)^2+a(b+c)^2+b(a+c)^2
    =c(a^2 + 2ab + b^2) + a(b^2 +2bc + c^2) + b(a^2 + 2ac + c^2)
    =ca^2 + 2abc + cb^2 + ab^2 + 2abc + ac^2 + ba^2 + 2abc + bc^2
    =a^2b + ac^2 + a^2c + b^2c + b^2a + bc^2 + 6abc

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