Rút gọn/ Thực hiện phép tính
a) (x - 2)(x^2 - 2x + 4)(x - 2)( x^2 + 2x + 4)
b) (a + b + c)^3 - (b + c - a)^3 - (a - b + c)^3 - (a + b - c)^3
c) (a + b)^3 + (b + c)^3 + (c + a)^3 - 3(a + b)(b + c)(c + a)
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a) (x - 2)(x2 - 2x + 4)(x - 2)( x2 + 2x + 4)
= (x - 2)2(x - 2)2(x + 2)2
= (x - 2)4(x + 2)2
b) (a + b + c)3 - (b + c - a)3 - (a - b + c)3 - (a + b - c)3
Đặt a+b-c=x, c+a-b=y, b+c-a=z
=>x+y+z=a+b-c+c+a-b+b+c-a=a+b+c
Ta có hằng đẳng thức:
(x+y+z)^3-3x-3y-3z=3(x+y)(x+z)(y+z)
=>(a+b+c)^3-(b+c-a)^3-(a+c-b)^3-(a+b-c)^3=(x+y+z)^3-x^3-y^3-z^3
=3(x+y)(x+z)(y+z)
=3(a+b-c+c+a-b)(c+a-b+b+c-a)(b+c-a+a+b-c)
=3.2a.2b.2c
=24abc
c) (a + b)3 + (b + c)3 + (c + a)3 - 3(a + b)(b + c)(c + a)
Đặt x = a+b; y = b+c; z = c+a ta có:
x3+y3+z3−3xyz
= (x+y)3−3xy(x−y)+z3−3xyz
=[(x+y)3+z3]−3xy(x+y+z)
=(x+y+z)3−3z(x+y)(x+y+z)−3xy(x−y−z)
=(x+y+z)[(x+y+z)2−3z(x+y)−3xy]
=(x+y+z)(x2+y2+z2+2xy+2xz+2yz−3xz−3yz−3xy)
=(x+y+z)(x2+y2+z2−xy−yz−yx)
Thay vào ta có:
(a+b+b+c+c+a)[(a+b)2+(b+c)2+(c+a)2−(a+b)(b+c)−(b+c)(c+a)−(c+a)(a+b)]
=(2a+2b+2c)(a2−ab−ac+b2−bc+c2)
=2(a+b+c)(a2−ab−ac+b2−bc+c2)
Bài 2:
a: =a-b+c+a-c+b-b
=2a-b
b: =2x-5+x-a+x-5-a
=4x-10-2a
a) (2x+3)2-2(2x+3)(2x+5)+(2x+5)2
=4x2+12x+9-(4x+6)(2x+5)+4x2+20x+25
=4x2+12x+9-(8x2+12x+20x+30)+4x2+20x+25
=4x2+12x+9-8x2-12x-20x-30+4x2+20x+25
=4
b) (x2+x+1)(x2-x+1)(x2-1)
=((x2+1)2-x2)(x2-1)
=(x4+x2+1)(x2-1)
=x6+x4+x2-x4-x2-1
=x6-1
c)(a+b-c)2+(a-b+c)2-2(b-c)2
=a2+b2+c2+2ab-2ac-2bc+a2+b2+c2-2ab+2ac-2bc-2(b2-2bc+c2)
=2a2+2b2+2c2-4bc-2b2+4bc-2c2
=2a2
d) (a+b+c)2+(a-b-c)2+(b-c-a)2+(c-a-b)2
= a2+b2+c2+2ab+2ac+2bc+a2+b2+c2-2ab-2ac+2bc+a2+b2+c2+2bc-2ab+2ac+a2+b2+c2-2ac-2bc+2ab
=4a2+4b2+4c2+4ab+4bc
a)\(\left(x-2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x-4\right)\)
\(=\left(x-2\right)^2\left(x^2-2x+4\right)\left(x^2+2x-4\right)\)
\(=\left(x-2\right)^2\left(x^4+4x^2+16\right)\)
\(=x^6-4x^5+8x^4-16x^3+32x^2-64x+64\)