作业帮多项式除法 是怎么运算的!
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/07/06 00:24:04
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?
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?
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)
再问： 朋友 不是让你反向推，不是让你同分的。 是让你分解的。
再答： (ax+b)/(cx+d)=[acx+ad)+(bc-ad)]/[c(cx+d)]=[a(cx+d)+ (bc-ad)]/[c(cx+d)]=a/c+(bc-ad)/[c(cx+d)] =a/c+(bc-ad)/(c²)*1/(x+(d/c)) (x+3)+(7x-6)/( (x-2)(x-1) )=x+3+[8(x-1)-(x-2)]/((x-2)(x-1) )=x+3+8/(x-2)-1/(x-1) x³/(x²-3x+2)= [x^3-3x^2+2x+3x^2-9x+6+7x-6]/(x²-3x+2) =[(x+3)(x^2-3x+2)+7x-6]/(x²-3x+2)=[(x+3)(x-2)(x-1)+(7x-6)]/((x-2)(x-1) )=(x+3)+(7x-6)/((x-2)(x-1) )
=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)
再问： 朋友 不是让你反向推，不是让你同分的。 是让你分解的。
再答： (ax+b)/(cx+d)=[acx+ad)+(bc-ad)]/[c(cx+d)]=[a(cx+d)+ (bc-ad)]/[c(cx+d)]=a/c+(bc-ad)/[c(cx+d)] =a/c+(bc-ad)/(c²)*1/(x+(d/c)) (x+3)+(7x-6)/( (x-2)(x-1) )=x+3+[8(x-1)-(x-2)]/((x-2)(x-1) )=x+3+8/(x-2)-1/(x-1) x³/(x²-3x+2)= [x^3-3x^2+2x+3x^2-9x+6+7x-6]/(x²-3x+2) =[(x+3)(x^2-3x+2)+7x-6]/(x²-3x+2)=[(x+3)(x-2)(x-1)+(7x-6)]/((x-2)(x-1) )=(x+3)+(7x-6)/((x-2)(x-1) )