Phần tích đa thức thành nhân tử a, (x+1)(x+2)(x+3)(x+4)-24 b, (x^2+2x)^2+9x^2+18x+20 c, (a+b+c)^3-(a+b-c)^3-(b+c-a)^3-(c+a-b)^3

Đáp án:

a, Ta có :

`A = (x + 1)(x + 2)(x + 3)(x + 4) – 24`

` = [(x + 1)(x + 4)].[(x + 2)(x + 3)] – 24`

`= (x^2 + x + 4x + 4)(x^2 + 2x + 3x + 6) – 24`

`= (x^2 + 5x + 4)(x^2 + 5x + 6) – 24`

Dặt `t = x^2 + 5x + 5`

`=> A = (t – 1)(t + 1) – 24`

`= t^2 – 1 – 24`

`= t^2 – 25`

`= t^2 – 5^2`

`= (t – 5)(t + 5)`

`=> A = (x^2 + 5x + 5 – 5)(x^2 + 5x + 5 + 5)`

`= (x^2 + 5x)(x^2 + 5x + 10)`

b, Ta có : 

`(x^2 + 2x)^2 + 9x^2 + 18x + 20`

`= (x^2 + 2x)^2 + 9(x^2 + 2x) + 20`

`= (x^2 + 2x)^2 + 2.(x^2 + 2x) . 9/2 + 81/4 – 1/4`

`= (x^2 + 2x + 9/2)^2 – (1/2)^2`

`= (x^2 + 2x + 9/2 – 1/2)(x^2 + 2x + 9/2 + 1/2)`

`= (x^2 + 2x + 4)(x^2 + 2x + 5)`

c, Ta có : 

`B = (a + b + c)^3 – (a + b – c)^3 – (b + c – a)^3 – (c + a – b)^3`

`= [(a + b + c)^3 – (a + b – c)^3] – [(b + c – a)^3 + (c + a – b)^3]`

` = (a + b + c – a – b + c)^3 + 3(a + b + c)(a + b – c)[(a + b + c) – (a + b – c)] – [(b + c – a + c + a – b)^3 – 3(b + c – a)(c + a – b)(b + c – a + c + a – b)`

`= (2c)^3 + 3(a + b + c)(a + b – c)(a + b + c – a – b + c) – (2c)^3 + 3(b + c – a)(c + a – b)(b + c – a + c + a – b)`

`= 3(a + b + c)(a + b – c).2c  + 3(b + c – a)(c + a – b).2c`

`= 6c.[(a + b + c)(a + b – c) + (b + c – a)(a + c – b)]`

Đặt `b + c – a = x` ; `c + a – b = y` ; `a + b – c = z`

`=> x + y + z = a + b + c`

`=> B = 6c.[(x + y + z).z + x.y]`

` = 6c(zx + zy + z^2 + xy)`

`= 6c[(zx + xy) + (zy + z^2)]`

`= 6c[x(z + y) + z(y + z)]`

`= 6c(x + z)(y + z)`

`= 6c(b + c – a + a + b – c)(c + a – b + a + b  – c)`

`= 6c.2b.2a`

`= 24abc`

Giải thích các bước giải:

 

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