Toán 8 Rút gọn các biểu thức

Darkness Evolution

Duke of Mathematics
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Vĩnh Phúc
THCS Vĩnh Yên
Ghép 2 đứa đầu với nhau, 2 đứa cuối với nhau
(a+b+c)3(b+ca)3(c+ab)3(a+bc)3(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3
=(a+b+c)3(b+ca)3[(c+ab)3+(a+bc)3]= (a+b+c)^3-(b+c-a)^3-[(c+a-b)^3+(a+b-c)^3]
=(a+b+cbc+a)[(a+b+c)2+(a+b+c)(b+ca)+(b+ca)2](c+ab+a+bc)[(c+ab)2(c+ab)(a+bc)+(a+bc)2]=(a+b+c-b-c+a)[(a+b+c)^2+(a+b+c)(b+c-a)+(b+c-a)^2]-(c+a-b+a+b-c)[(c+a-b)^2-(c+a-b)(a+b-c)+(a+b-c)^2]
=2a[(a+b+c+b+ca)2(a+b+c)(b+ca)2a[(c+abab+c)2+(c+ab)(a+bc)]=2a[(a+b+c+b+c-a)^2-(a+b+c)(b+c-a)-2a[(c+a-b-a-b+c)^2+(c+a-b)(a+b-c)]
=2a[4(b+c)2((b+c)2a2)4(cb)2((a2(bc)2)]=2a[4(b+c)^2-((b+c)^2-a^2)-4(c-b)^2-((a^2-(b-c)^2)]
=2a[3(b+c)2+a2a23(bc)2]=2a[3(b+c)^2+a^2-a^2-3(b-c)^2]
=2a12bc=24abc=2a \cdot 12bc =24abc
(Còn cách nữa, edit sau)
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Cách 2: Đặt b+ca=x;c+ab=y;a+bx=zb+c-a=x; c+a-b=y; a+b-x=z
x+y+z=a+b+c\Rightarrow x+y+z=a+b+c
Khi đó, biểu thức trở thành (x+y+z)3x3y3z3(x+y+z)^3-x^3-y^3-z^3
=(x+y+z)3x3(y3+z3)=(x+y+z)^3-x^3-(y^3+z^3)
=(y+z)[(x+y+z)2+x(x+y+z)+x2)(y+z)(y2yz+z2)=(y+z)[(x+y+z)^2+x(x+y+z)+x^2)-(y+z)(y^2-yz+z^2)
=(y+z)(x2+y2+z2+2xy+2yz+2zx+x2+xy+xz+x2y2+yzz2)=(y+z)(x^2+y^2+z^2+2xy+2yz+2zx+x^2+xy+xz+x^2-y^2+yz-z^2)
=(y+z)(3x2+3xy+3xz+3yz)=(y+z)(3x^2+3xy+3xz+3yz)
=3(y+z)(x+y)(x+z)=32a2b2c=24abc=3(y+z)(x+y)(x+z)=3 \cdot 2a \cdot 2b \cdot 2c =24abc
 
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Duy Quang Vũ 2007

Học sinh chăm học
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THCS Chu Văn An
Đặt x=a+bc,y=b+ca,z=c+abx+y+z=a+b+cx=a+b-c,y=b+c-a,z=c+a-b \Rightarrow x+y+z=a+b+c
Do đó: (a+b+c)3(a+bc)3(b+ca)3(c+ab)3=(x+y+z)3x3y3z3=[(x+y+z)x][(x+y+z)2+(x+y+z)x+x2](y+z)(y2yz+z2)=(y+z)(x2+y2+z2+2xy+2yz+2zx+x2+yx+zx+x2)(y+z)(y2yz+z2)=(y+z)(x2+y2+z2+2xy+2yz+2zx+x2+yx+zx+x2y2+yzz2)=(y+z)(3x2+3xy+3yz+3zx)=3(y+z)(x2+xy+yz+zx)=3(y+z)[x(x+y)+z(x+y)]=3(x+y)(y+z)(z+x)(a+b+c)^3-(a+b-c)^3-(b+c-a)^3-(c+a-b)^3\\ =(x+y+z)^3-x^3-y^3-z^3\\ =[(x+y+z)-x][(x+y+z)^2+(x+y+z)x+x^2]-(y+z)(y^2-yz+z^2)\\ =(y+z)(x^2+y^2+z^2+2xy+2yz+2zx+x^2+yx+zx+x^2)-(y+z)(y^2-yz+z^2)\\ =(y+z)(x^2+y^2+z^2+2xy+2yz+2zx+x^2+yx+zx+x^2-y^2+yz-z^2)\\ =(y+z)(3x^2+3xy+3yz+3zx)\\ =3(y+z)(x^2+xy+yz+zx)\\ =3(y+z)[x(x+y)+z(x+y)]\\ =3(x+y)(y+z)(z+x)
Ta có: x+y=(a+bc)+(b+ca)=2b;y+z=(b+ca)+(c+ab)=2c;z+x=(c+ab)+(a+bc)=2ax+y=(a+b-c)+(b+c-a)=2b;\\ y+z=(b+c-a)+(c+a-b)=2c;\\ z+x=(c+a-b)+(a+b-c)=2a
Do đó: (a+b+c)3+(a+bc)3+(b+ca)3+(c+ab)3=3.2a.2b.2c=24abc(a+b+c)^3+(a+b-c)^3+(b+c-a)^3+(c+a-b)^3=3.2a.2b.2c=24abc
 
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