多项式除法 是怎么运算的!y=(ax+b)/(cx+d) y=a/c+(bc-ad)/(c²) *1/(x+(d/c))这个是怎么算的啊 y=x³/(x²-3x+2)y=(x+3)+(7x-6)/( (x-2)(x-1) )=x+3+8/(x-2)-1/(x-1)怎么算的?y=(ax+b)/(cx+d) y=a/c+(bc-ad)/(c²) *1/(x+(d/c))
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多项式除法 是怎么运算的!y=(ax+b)/(cx+d) y=a/c+(bc-ad)/(c²) *1/(x+(d/c))这个是怎么算的啊 y=x³/(x²-3x+2)y=(x+3)+(7x-6)/( (x-2)(x-1) )=x+3+8/(x-2)-1/(x-1)怎么算的?y=(ax+b)/(cx+d) y=a/c+(bc-ad)/(c²) *1/(x+(d/c))
多项式除法 是怎么运算的!
y=(ax+b)/(cx+d) y=a/c+(bc-ad)/(c²) *1/(x+(d/c))这个是怎么算的啊 y=x³/(x²-3x+2)y=(x+3)+(7x-6)/( (x-2)(x-1) )=x+3+8/(x-2)-1/(x-1)怎么算的?
y=(ax+b)/(cx+d)
y=a/c+(bc-ad)/(c²) *1/(x+(d/c))
这个是怎么算的啊
y=x³/(x²-3x+2)
y=(x+3)+(7x-6)/( (x-2)(x-1) )=x+3+8/(x-2)-1/(x-1)
怎么算的?
多项式除法 是怎么运算的!y=(ax+b)/(cx+d) y=a/c+(bc-ad)/(c²) *1/(x+(d/c))这个是怎么算的啊 y=x³/(x²-3x+2)y=(x+3)+(7x-6)/( (x-2)(x-1) )=x+3+8/(x-2)-1/(x-1)怎么算的?y=(ax+b)/(cx+d) y=a/c+(bc-ad)/(c²) *1/(x+(d/c))
y=a/c+(bc-ad)/(c²) *1/(x+(d/c))
=a/c+ (bc-ad)/[c(cx+d)]=[a(cx+d) + (bc-ad)]/[c(cx+d)]=[acx+ad)+ (bc-ad)]/[c(cx+d)]
=(ax+b)/(cx+d)
x+3+8/(x-2)-1/(x-1)
=x+3+[8(x-1)-(x-2)]/( (x-2)(x-1) )=(x+3)+(7x-6)/( (x-2)(x-1) )
(x+3)+(7x-6)/( (x-2)(x-1) )=[(x+3)(x-2)(x-1)+(7x-6)]/( (x-2)(x-1) )
=[(x+3)(x^2-3x+2)+7x-6]/(x²-3x+2)
=[x^3-3x^2+2x+3x^2-9x+6+7x-6]/(x²-3x+2)
=x³/(x²-3x+2)