分式计算,下列计算请把计算过程详细地写出来,(1)2a/5a²b + 3b/10ab²(2)3y/2x+2y + 2xy/x²+xy (3)2x/x²-64y² - 1/x-8y
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![分式计算,下列计算请把计算过程详细地写出来,(1)2a/5a²b + 3b/10ab²(2)3y/2x+2y + 2xy/x²+xy (3)2x/x²-64y² - 1/x-8y](/uploads/image/z/4797714-66-4.jpg?t=%E5%88%86%E5%BC%8F%E8%AE%A1%E7%AE%97%2C%E4%B8%8B%E5%88%97%E8%AE%A1%E7%AE%97%E8%AF%B7%E6%8A%8A%E8%AE%A1%E7%AE%97%E8%BF%87%E7%A8%8B%E8%AF%A6%E7%BB%86%E5%9C%B0%E5%86%99%E5%87%BA%E6%9D%A5%2C%EF%BC%881%EF%BC%892a%2F5a%26sup2%3Bb+%2B+3b%2F10ab%26sup2%3B%EF%BC%882%EF%BC%893y%2F2x%2B2y+%2B+2xy%2Fx%26sup2%3B%2Bxy+%EF%BC%883%EF%BC%892x%2Fx%26sup2%3B-64y%26sup2%3B+-+1%2Fx-8y)
分式计算,下列计算请把计算过程详细地写出来,(1)2a/5a²b + 3b/10ab²(2)3y/2x+2y + 2xy/x²+xy (3)2x/x²-64y² - 1/x-8y
分式计算,下列计算请把计算过程详细地写出来,
(1)2a/5a²b + 3b/10ab²
(2)3y/2x+2y + 2xy/x²+xy
(3)2x/x²-64y² - 1/x-8y
分式计算,下列计算请把计算过程详细地写出来,(1)2a/5a²b + 3b/10ab²(2)3y/2x+2y + 2xy/x²+xy (3)2x/x²-64y² - 1/x-8y
(1) 2a/5a²b + 3b/10ab²
=2/(5ab) + 3/(10ab) 前项约a; 后项约b;
=4/(10ab) + 3/(10ab) 通分
=7/(10ab) 合并
(2)3y/2x+2y + 2xy/x²+xy
=3y/(2x+2y) + 2xy/[x(x+y)] 估计你原题是代括号的
=3y/[ 2(x+y) ] + 2xy / [x(x+y)] 前项分母提取公因数2; 后项分母提取公因数x;
=3y/[ 2(x+y) ] + 4y / [2(x+y)] 前项不变照写;后项分子分母同约x,同时后项分母配2;
=(3y+ 4y) /[ 2(x+y) ] 上面前后项已经通分,合并
=7y /[ 2(x+y) ] 继续合并分子
(3)2x/x²-64y² - 1/x-8y
=2x/(x²-64y²) - 1/(x-8y) 估计你原题是代括号的
=2x / [(x-8y) (x+8y)] - 1/(x-8y) 用到平方差公式:a^2-b^2=(a-b)(a+b)
=2x / [(x-8y) (x+8y)] - (x+8y) /([(x-8y) (x+8y)] 通分使前后两项分母相同,准备合并
=[2x-(x+8y) ] / [(x-8y) (x+8y)] 合并前后两项
=[2x-x-8y ] / [(x-8y) (x+8y)] 注意:a-(b+c) =a-b-c,而不是=a-b+c
=[x-8y ] / [(x-8y) (x+8y)] 约分化简
=1 / (x+8y) Game over !§ ※
补充一条:分数的分母是不能为零滴,虽然这几个题目没涉及到.