math -factorisation

2007-05-12 5:13 pm
factorise
(1)a^2(b+c)+b^2(c-a)+c^2(b-a)-2abc

(2)x^2(b-c)+b^2(c+x)-c^2(x+b)

(3)x^2(b^2-c^2)+b(c^2-x^2)+c(x^2-b^2)

(4)x^2(y+z)+y^2(z-x)+z^2(y-x)-2xyz

回答 (2)

2007-05-12 9:00 pm
✔ 最佳答案
(1) a²(b + c) + b²(c - a) + c²(b - a) - 2abc
= a²b + a²c + b²c - ab² + bc² - ac² - 2abc
= a²b - 2abc + bc² + a²c - ac² - ab² + b²c
= b(a² - 2ac + c²) + ac(a - c) - b²(a - c)
= b(a - c)² + ac(a - c) - b²(a - c)
= (a - c)[ b(a - c) + ac - b² ]
= (a - c)(ab - bc + ac - b²)
= (a - c)(ab - b² + ac - bc)
= (a - c)[ b(a - b) + c(a - b)]
= (a - c)(a - b)(b + c)

(2) x²(b - c) + b²(c + x) - c²(x + b)
= x²(b - c) + b²c + b²x - c²x - bc²
= x²(b - c) + b²c - bc² + b²x - c²x
= x²(b - c) + bc(b -c) + x (b² - c²)
= x²(b - c) + bc(b -c) + x (b - c)(b + c)
= (b - c)[x² + bc + x(b+c)]
= (b - c)(x² + bc + bx + cx)
= (b - c)(cx + x² + bc + bx)
= (b - c)[x(c + x) + b(c + x)]
= (b - c)(c + x)( b + x)

(3) x²(b²-c²)+b(c²-x²)+c(x²-b²)
= x²(b + c)(b - c) + bc² - bx² + cx² - b²c
= x²(b + c)(b - c) - bx² + cx² - b²c + bc²
= x²(b + c)(b - c) - x² (b - c) - bc(b - c)
= (b - c) [x²(b + c) - x² - bc]
= (b - c) [x²(b + c -1) - bc]

(4) x²(y+z)+y²(z-x)+z²(y-x)-2xyz
= x²y + x²z + y²z - xy² + yz² - xz² - 2xyz
= x²y - 2xyz + yz² + x²z - xz² - xy² + y²z
= y(x² - 2xz + z²) + xz(x - z) - y²(x - z)
= y(x - z)² + xz(x - z) - y²(x - z)
= (x - z)[ y(x - z) + xz - y² ]
= (x - z)(xy - yz + xz - y²)
= (x - z)(xy - y² + xz - yz)
= (x - z)[ y(x - y) + z(x - y)]
= (x - z)(x - y)(y + z)
2007-05-12 8:49 pm
2)
(b-c)x^2 + (b^2 - c^2)x + (b^2 c - b c^2)
= (b-c)x^2 + (b-c)(b+c)x + bc(b-c)
= (b-c)[x^2 + (b+c)x + bc]
= (b-c)(x+b)(x+c)

做住一條,唔得閒..


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