添加和减去指数
在数学中,指数用于以幂表示更大的数字。它表示基数与自身相乘的次数。基数是任何数字或任何数学表达式。例如,此处的 A 3底数为 A,幂为 3,这意味着 A 将自身相乘 3 次,即 A 3 = A x A x A。指数的一般项为
Yn = Y × Y × Y ×………n times
这里,y 被称为底数,n 被称为幂或指数。指数类型:
- 负指数:负指数是那些指数,它告诉多少倍的基础倍数与自身的倒数。它表示为-n或 1/a n 。例如, 23 -2 、 4 -2 。
- 分数指数:当指数以分数表示时,这种类型的指数称为分数指数。它表示为1/n 。例如, 3 1/2 、 4 1/3 。
- 十进制指数:当指数以十进制数字表示时,这种类型的指数称为十进制指数。它表示为1.3 。例如, 3 1.5 、 4 12.3 。
添加和减去指数
加法和减法是数学的两个基本运算。加法表示求两位数之和,减法表示求两位数之差。但不能直接加减指数,只能对同底同幂的系数或变量进行加减。我们只能在乘法中加指数,在除法中减指数。
在代数中的指数之间进行加法或减法的步骤:
To do addition or subtraction we will follow the same steps:
Step 1: Before performing any addition or subtraction between exponents we need to observe whether the base and exponents are the same or not.
Step 2: Arrange the similar variables/terms together.
Step 3: Now perform addition or subtraction as per need between the coefficient of terms.
示例 1:求解 6x 3 + 12x 3 。
解决方案:
Here the base is same i.e., x and exponents of two terms are also same i.e., 3
So we can add coefficients of the two terms to get result.
6x3 + 12x3 = (6 + 12)x3
= 18x3
示例 2:求解 9x 3 -13x 3 。
解决方案:
Here the base is same i.e., x and exponents of two terms are also same i.e., 3
So we can subtract coefficients of the two terms to get result.
9x3 – 13x3 = (9 – 13)x3
= -4x3
在数字中的指数之间进行加法或减法的步骤:
To do addition or subtraction we will follow the same steps:
Step 1: Before performing any addition or subtraction between exponents we need to observe whether the base and exponents are the same or not.
Step 2: Arrange the similar base and exponent terms together. If the terms have different base and exponent then solve them individually.
Step 3: Now perform addition or subtraction as per need between the base of terms.
示例 1:求解 6 3 + 6 3 。
解决方案:
Here the base is same i.e., 6 and exponents of two terms are also same i.e., 3
So, we are solving them together
63 + 63 = 2(6)3
= 2 x 6 x 6 x 6
= 432
示例 2:求解 9 2 – 13 3 。
解决方案:
Here the base and the exponents are different
So, we are solving them individually.
92 – 133 = 9 x 9 -13 x 13 x 13
= 81 – 2197
= -2116
类似问题
问题 1:求解 5x 3 + 3x 3 。
解决方案:
Here the base is same i.e., x and exponents of two terms are also same i.e., 3
So we can add coefficients of the two terms to get result.
5x3 + 3x3 = (5 + 3)x3
= 8x3
So, 5x3 + 3x3 = 8x3
问题 2:表达式 -11a 2 + 4a 2的结果是什么。
解决方案:
Here the base is same i.e., a and exponents of two terms are also same i.e., 2
So we can add coefficients of the two terms to get result.
-11a2 + 4a2 = (-11 + 4)a2
= (4 – 11)a2
= -7a2
So, -11a2 + 4a2 = -7a2
问题 3:求解表达式 4x 3 + 4x 2 – 2x 3 + x 2 – x + 1
解决方案:
Here we have different kind of terms (x3, x2, x)i.e., bases are same but different exponents.
So identify the similar kind of variables and group them and perform addition/subtraction based on signs and arrange in polynomial order i.e., bases having higher exponential at first and lower at last.
4x3 + 4x2 – 2x3 + x2 – x + 1 = (4x3 – 2x3) + (4x2 + x2)-x + 1
= (4 – 2)x3 + (4 + 1)x2 – x + 1
= 2x3 + 5x2 – x + 1
问题 4:x 3 y + 4x 3 y 的结果是什么?
解决方案:
Here we have 2 different variables in 2 terms x, y and the exponents of x, y in two terms are same i.e., 3,1 respectively.
So we can consider that this 2 terms has matching variables and we can add/subtract the coefficient of 2 terms based on requirement.
x3y + 4x3y = (1 + 4)x3y
= 5x3y
问题 5:求解 x 3 y + 4x 3 y 2 + 4x – x + 1
解决方案:
Here we have 2 different variables in 2 terms x, y and the exponent of x in 2 terms are same i.e., 3 but the exponent of y is not same so we can’t consider the 2 terms as same and we won’t perform any operation between them. (remained as it is)
But there are other two more terms with same base variable x and exponent as 1. So we group them and perform computations on its coefficients.
x3y + 4x3y2 + 4x – x + 1 = 4x3y2 + x3y + (4x – x) + 1
= 4x3y2 + x3y + (4 – 1)x + 1
= 4x3y2 + x3y + 3x + 1
问题 6:求解 x 5 y 2 – x 4 y 4
解决方案:
Here there are two terms x5y2, x4y4 and in two terms we have 2 variables x, y but the exponents of variables are not same when made comparison between terms.
Hence the first term is entirely not like the second term and can’t be subtracted from each other.
We leave them as it is.
x5y2 – x4y4 can’t be simplified further.