代数恒等式是一个等式,适用于其变量的任何值。它们通常用于多项式的因式分解或代数计算的简化。许多日常情况都可以用数学方程式来表述。身份通过给予因素分解来帮助我们。
在操作手机时。手机屏幕下方有数百万个芯片,它们工作出色,因此您几乎不会感到任何毛刺。成千上万的工程师就复杂的方程式集思广益,使之成为可能。他们处理更大的方程式和问题。代数身份然后成为帮助解决这些问题的小工具。因此,也许它们不是直接出现在我们面前,而是在幕后的某个地方,肯定有人使用了它们,并使我们的生活更加轻松。
多项式的类型
基于代数表达式中存在的项数,例如,可以有1个项,2个项,3个项,4个项,依此类推,将它们分为不同的类别。术语之间用正号或负号分隔:
Types of Polynomials | Definition | Example |
Monomials | A Polynomial containing only one term |
10, 2x2, 4abc |
Binomials | A Polynomial containing two terms |
x+y, 3p2-5, x3y+8z |
Trinomials | A Polynomial containing three terms |
p+q+r, x5+5x+3, |
Quadrinomial | A Polynomial containing four terms |
p+q+r+s, m2n-2mn+3m+7 |
Quintinomial | A Polynomial containing five terms | 5x3y+ 6x2– 9xy+8y-7 |
Note:
- Quadrinomials and Quintinomials are not very famous terms and are often referred as Polynomials.
- The most common Standard Identities are from Binomials and Trinomials.
代数表达的标准恒等式
所有标准代数恒等式均来自二项式定理。有许多代数恒等式,但下面列出了一些标准的恒等式:
Standard Identities |
(a + b)2 = a2 + b2 + 2ab |
(a – b)2 = a2 + b2 -2ab |
a2 – b2 = (a + b)(a – b) |
(ax + b)(ax – b) = ax2 – b2 |
(x + a) (x + b) = x2 + (a + b)x + ab |
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) |
让我们看一下这些身份的不同应用。
身份申请
身份1:(a + b) 2 = a 2 + b 2 + 2ab
问题:当x = 3时,使用代数恒等式求出(x + 6)(x + 6)的值。
解决方案:
(x+6)(x+6) can be re-written as (x + 6)2.
It can be rewritten in this form, (a + b)2 = a2 + b2 + 2ab.
(x + 6)2 = x2 + 62 + 2(6x)
= x2 + 36 + 12x
Given, x = 3.
(x + 6)2 = 32 + 36 + 12(3)
= 9 + 36 + 36
= 81
身份2:(a – b) 2 = a 2 + b 2 -2ab
问题:展开(5x – 3y) 2 。
解决方案:
This is similar to expanding (a – b)2 = a2 + b2 – 2ab.
where a = 5x and b = 3y,
So (5x – 3y)2 = (5x)2 + (3y)2 – 2(5x)(3y)
= 25x2 + 9y2 – 30xy
身份3:a 2 – b 2 =(a + b)(a – b)
问题:使用上述身份分解(x 6 – 1)。
解决方案:
(x6 – 1) can be written as (x3)2 – 12.
This resembles the identity a2 – b2 = (a + b)(a – b).
where a = x3, and b = 1.
So, x6 – 1 = (x3)2 – 1 = (x3 + 1) (x3 – 1).
其他一些问题和应用
问题1:如果a + b = 12且ab = 35,那么a 4 + b 4是什么?
解决方案:
a4 + b4 can be written as (a2)2 + (b2)2,
and we know, (x + y)2 = x2 + y2 + 2xy
⇒ x2 + y2 = (x + y)2 -2xy
So, in this case, x = a2, y = b2 ;
a4 + b4 = (a2 + b2)2 -2(a2)(b2)
⇒ ((a+b)2 – 2ab)2 – 2(a2)(b2)
⇒ ((12)2 – 2(35))2 – 2(35)2
⇒ 5475 – 2450
⇒ 3026
问题2:对于z的所有实数值,恒等式4(z + 7)(2z-1)= Az 2 + Bz + C成立。什么是A + B + C?
解决方案:
Multiplying out the left side of the identity, we have
4(x+7)(2x−1)=8x2 +52x−28.
This expression must be equal to the right-hand side of the identity, implying
8x2+52x-28=Ax2+Bx+C,
So now comparing both sides of the equation.
A = 8, B = 52 ad C -28.
A + B + C = 8 + 52 -28 = 32
问题3:如果a + b + c = 6,则2 + b 2 + c 2 = 14,而ab + bc + ca = 11是3 + b 3 + c 3 -3abc?
解决方案:
We know this identity,
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab -bc -ca)
Substituting the given values,
a3 + b3 + c3 -3abc = (6)(14 -11)
⇒ (6)(3) = 18