4道计算题
1》[(4x-3y)/(3xy²z)]-[(2y-5z)/(3xy²z)]-[(4x+5z)/(3xy²z)]2》[3/(x-4)]-[24...
1》[(4x-3y)/(3xy²z)]-[(2y-5z)/(3xy²z)]-[(4x+5z)/(3xy²z)]
2》[3/(x-4)]-[24/(x²-16)]
3》[(a²+b²)/(a-b)]-a-b
4》[7/(2x²+6x)]-[(11-2x²)/(9-x²)]+2 展开
2》[3/(x-4)]-[24/(x²-16)]
3》[(a²+b²)/(a-b)]-a-b
4》[7/(2x²+6x)]-[(11-2x²)/(9-x²)]+2 展开
1个回答
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1)原式=[(4x-3y)—(2y-5z)—(4x+5z)]/3xy²z=-5y/3xy²z=-5/3xyz
2)原式=[3/(x-4)]-[24/(x+4)(x-4)]=[3(x+4)/(x+4)(x-4)]-[24/(x+4)(x-4)]=[3(x+4)-24]/[(x+4)(x-4)]=(3x-12)/[(x+4)(x-4)]=3/(x+4)
3)原式=[(a²+b²)/(a-b)]-(a+b)=[(a²+b²)/(a-b)]-[(a+b)(a-b)/(a-b)]=[(a²+b²)/(a-b)]-[(a²-b²)/(a-b)]=2b²/(a-b)
4)原式=[7/2x(x+3)]-[(11-2x²)/(3+x)(3-x)]+2
以下先把分母统一成2x(3+x)(3-x)然后计算
这类题目解法都一样,都是先统一分母,然后计算合并,希望对你有帮助
2)原式=[3/(x-4)]-[24/(x+4)(x-4)]=[3(x+4)/(x+4)(x-4)]-[24/(x+4)(x-4)]=[3(x+4)-24]/[(x+4)(x-4)]=(3x-12)/[(x+4)(x-4)]=3/(x+4)
3)原式=[(a²+b²)/(a-b)]-(a+b)=[(a²+b²)/(a-b)]-[(a+b)(a-b)/(a-b)]=[(a²+b²)/(a-b)]-[(a²-b²)/(a-b)]=2b²/(a-b)
4)原式=[7/2x(x+3)]-[(11-2x²)/(3+x)(3-x)]+2
以下先把分母统一成2x(3+x)(3-x)然后计算
这类题目解法都一样,都是先统一分母,然后计算合并,希望对你有帮助
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