为什么要合理化分母?
代数是数学研究的一个广阔领域,与数论、算术、几何及其分析相关。代数与符号的研究及其与数学运算的操作有关。给定的文章是对分母和分子、合理化的研究,并解释了为什么要合理化分母。
分子和分母
分子是分数的顶部。它解释了给定分数中存在的部分对象的计数。分子一词来源于拉丁语“enumerate”,意思是“计数”。举个例子,分数 2/5 的分子说明给定对象被分成 5 个相等的部分,分数包含其中的两个。
分母是分数的底部。它解释了整个对象的多少部分被分解。术语分母本身来源于拉丁语“nomen”。分母表示所描述分子的分数类型。作为分数的分母的分母的一个例子是,比如说,5,那么这表明整个对象被分成 5 个相等的部分。
合理化
合理化是通过将一个 surd 与一个相似的 surd 相乘而获得一个有理数的过程。另一个成倍增加的术语是合理化因素(RF)。整个合理化过程是通过将平方根或立方根从分母移动到分子来进行的。
例如,使表达式 x + √y 合理化
合理化因素 x – √y
现在,
= (x + √y)(x – √y) = x 2 – (√y) 2
= x 2 – y
为什么要合理化分母?
回答:
While performing a basic operation we rationalize a denominator to get the calculation easier and obtain a rational number as a result. In the process of rationalization, we exclude the square roots, cube roots, or any other radical expressions from the equation. Let’s see the method of rationalization by an example.
Rationalize the expression (√3 – 1)/(√3 + 1)
In the expression, rationalizing factor of denominator that would be √3 – 1. Multiplying and dividing the rationalizing factor of denominator,
= (√3 – 1)/(√3 + 1) × (√3 – 1)/ (√3 – 1)
= (√3 – 1)2/(√3)2 – 1
By the formula (a – b)2 = a2 – 2ab + b2
= (√3)2 – 2√3 × 1 + (1)2/ (3 – 1)
= 4 – 2√3/2
Taking 2 in common in numerator
= 2(2 – √3)/2
Cancelling common factors
= 2 – √3
示例问题
问题一:合理化 2√3/√3
解决方案:
To rationalize the expression 2√3/√3 a rationalizing factor is needed which is √3.
Now,
= 2√3/√3 × √3/√3
= 2 × 3 /3
= 2
问题 2:合理化 (2 + √3)/√3
解决方案:
To rationalize the expression, 2 + √3/√3 we need a rationalizing factor which is √3.
= 2 + √3/√3 × √3/√3
= 2√3 + 3/3
问题 3:合理化 1/√x
解决方案:
To rationalize the expression 1/√x, rationalizing factor is required which is √x.
= 1/(√x x) √x/√x.
= √x /x
问题 4:合理化表达 32/5 – √7
解决方案:
32/5 – √7 to rationalise this expression, rationalizing factor is needed which is 5 + √7
= 32/5 – √7 × 5 + √7/5 + √7
= 32(5 + √7)/(5 – √7)(5 + √7)
= (160 – 32√7)/(25 + 5√7 – 5√7 – 7)
= (160 – 32√7)/32
= 5 – √7
问题 5:合理化分母 5 – √3/2 + √3
解决方案:
To rationalize the expression 5 – √3/2 + √3 a rationalizing factor is needed 2 – √3.
= 5 – √3/2 + √3 × 2 – √3/ 2 – √3
= (5 – √3)(2 – √3)/ (2)2 – (√3)2
= 10 – 5√3 – 2√3 + 3/4 – 3
= 13 – 7√3/1
= 13 – 7√3
问题 6:合理化 (√3 – 1)/(√3 + 1)。
解决方案:
To rationalize the expression (√3 – 1)/(√3 + 1) a rationalizing factor is needed √3 – 1
= √3 – 1/√3 + 1 × √3 – 1/√3 – 1
= (√3 – 1)2/(√3)2 – (1)2
= (3 + 1 – 2√3)/(3 – 1)
= (4 – 2√3)/2
= 2 – √3