Cho `a^3+b^3+c^3=3abc` Tính: `A=(1+a/b)(1+b/c)(1+c/a)` 02/07/2021 Bởi Ruby Cho `a^3+b^3+c^3=3abc` Tính: `A=(1+a/b)(1+b/c)(1+c/a)`
`a^3 + b^3 + c^3 = 3abc` `<=> a^3 + b^3 + c^3 – 3abc = 0` `<=> (a + b)^3 – 3ab(a + b) + c^3 – 3abc = 0` `<=> [(a + b)^3 + c^3] – [3ab(a + b) + 3abc] = 0` `<=> (a + b + c)[(a + b)^2 – (a + b)c + c^2] – 3ab(a + b + c) = 0` `<=> (a + b + c)(a^2 + 2ab + b^2 – ac – bc + c^2 – 3ab) = 0` `<=> (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca) = 0` `+)a + b + c = 0` `=> a + b = -c` `b + c = -a` `c + a = -b` `+)A = (1 + a/b)(1 + b/c)(1 + c/a)` `= (a + b)/b + (b + c)/c + (c + a)/a` `= [(a + b)(b + c)(c + a)]/(abc)` `= [(-c).(-a).(-b)]/(abc)` `= (-abc)/(abc)` `= -1` `+)a^2 + b^2 + c^2 – ab – bc – ca = 0` `<=> 2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca = 0` `<=> (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 0` `<=> (a – b)^2 + (b – c)^2 + (c – a)^2 = 0` `<=> a = b = c` `=> M = (1 + a/b)(1 + b/c)(1 + c/a)` `= (1 + a/a)(1 + b/b)(1 + c/c)` `= (1 + 1)(1 + 1)(1 + 1)` `=8` Bình luận
Đáp án: Ta có : `a^3 + b^3 + c^3 = 3abc` `<=> a^3 + b^3 + c^3 – 3abc = 0` `<=> (a + b)^3 – 3ab(a + b) + c^3 – 3abc = 0` `<=> [(a + b)^3 + c^3] – [3ab(a + b) + 3abc] = 0` `<=> (a + b + c)[(a + b)^2 – (a + b)c + c^2] – 3ab(a + b + c) = 0` `<=> (a + b + c)(a^2 + 2ab + b^2 – ac – bc + c^2 – 3ab) = 0` `<=> (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca) = 0` TH1 : `a + b + c = 0` `=> a + b = -c` `b + c = -a` `c + a = -b` Ta có : `A = (1 + a/b)(1 + b/c)(1 + c/a)` `= (a + b)/b + (b + c)/c + (c + a)/a` `= [(a + b)(b + c)(c + a)]/(abc)` `= [(-c).(-a).(-b)]/(abc)` `= (-abc)/(abc)` `= -1` th2 : `a^2 + b^2 + c^2 – ab – bc – ca = 0` `=> 2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca = 0` `=> (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 0` `=> (a – b)^2 + (b – c)^2 + (c – a)^2 = 0` `=> a – b = b – c = c – a = 0` `=> a = b = c` `=> M = (1 + a/b)(1 + b/c)(1 + c/a)` `= (1 + a/a)(1 + b/b)(1 + c/c)` `= (1 + 1)(1 + 1)(1 + 1)` `= 2 . 2 . 2` `=8` Giải thích các bước giải: Bình luận
`a^3 + b^3 + c^3 = 3abc`
`<=> a^3 + b^3 + c^3 – 3abc = 0`
`<=> (a + b)^3 – 3ab(a + b) + c^3 – 3abc = 0`
`<=> [(a + b)^3 + c^3] – [3ab(a + b) + 3abc] = 0`
`<=> (a + b + c)[(a + b)^2 – (a + b)c + c^2] – 3ab(a + b + c) = 0`
`<=> (a + b + c)(a^2 + 2ab + b^2 – ac – bc + c^2 – 3ab) = 0`
`<=> (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca) = 0`
`+)a + b + c = 0`
`=> a + b = -c`
`b + c = -a`
`c + a = -b`
`+)A = (1 + a/b)(1 + b/c)(1 + c/a)`
`= (a + b)/b + (b + c)/c + (c + a)/a`
`= [(a + b)(b + c)(c + a)]/(abc)`
`= [(-c).(-a).(-b)]/(abc)`
`= (-abc)/(abc)`
`= -1`
`+)a^2 + b^2 + c^2 – ab – bc – ca = 0`
`<=> 2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca = 0`
`<=> (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 0`
`<=> (a – b)^2 + (b – c)^2 + (c – a)^2 = 0`
`<=> a = b = c`
`=> M = (1 + a/b)(1 + b/c)(1 + c/a)`
`= (1 + a/a)(1 + b/b)(1 + c/c)`
`= (1 + 1)(1 + 1)(1 + 1)`
`=8`
Đáp án:
Ta có :
`a^3 + b^3 + c^3 = 3abc`
`<=> a^3 + b^3 + c^3 – 3abc = 0`
`<=> (a + b)^3 – 3ab(a + b) + c^3 – 3abc = 0`
`<=> [(a + b)^3 + c^3] – [3ab(a + b) + 3abc] = 0`
`<=> (a + b + c)[(a + b)^2 – (a + b)c + c^2] – 3ab(a + b + c) = 0`
`<=> (a + b + c)(a^2 + 2ab + b^2 – ac – bc + c^2 – 3ab) = 0`
`<=> (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca) = 0`
TH1 : `a + b + c = 0`
`=> a + b = -c`
`b + c = -a`
`c + a = -b`
Ta có :
`A = (1 + a/b)(1 + b/c)(1 + c/a)`
`= (a + b)/b + (b + c)/c + (c + a)/a`
`= [(a + b)(b + c)(c + a)]/(abc)`
`= [(-c).(-a).(-b)]/(abc)`
`= (-abc)/(abc)`
`= -1`
th2 : `a^2 + b^2 + c^2 – ab – bc – ca = 0`
`=> 2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca = 0`
`=> (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 0`
`=> (a – b)^2 + (b – c)^2 + (c – a)^2 = 0`
`=> a – b = b – c = c – a = 0`
`=> a = b = c`
`=> M = (1 + a/b)(1 + b/c)(1 + c/a)`
`= (1 + a/a)(1 + b/b)(1 + c/c)`
`= (1 + 1)(1 + 1)(1 + 1)`
`= 2 . 2 . 2`
`=8`
Giải thích các bước giải: