推荐回答(4个)
初中数学竞赛辅导资料(20)
代数恒等式的证明
甲内容提要
证明代数恒等式,在整式部分常用因式分解和乘法两种相反的恒等变形,要特别注意运用乘法公式和等式的运算法则、性质。
具体证法一般有如下几种
1.从左边证到右边或从右边证到左边,其原则是化繁为简。变形的过程中要不断注意结论的形式。
2.把左、右两边分别化简,使它们都等于第三个代数式。
3.证明:左边的代数式减去右边代数式的值等于零。即由左边-右边=0可得左边=右边。
4,由己知等式出发,经过恒等变形达到求证的结论。还可以把己知的条件代入求证的一边证它能达到另一边,
乙例题
例1求证:3 n+2-2 n+2+2×5 n+2+3 n-2 n=10(5 n+1+3 n-2 n-1)
证明:左边=2×5×5 n+1+(3 n+2+3 n)+(-2 n+2 -2 n)
=10×5 n+1+3 n(32+1)-2 n-1(23+2)
=10(5 n+1+3 n-2 n-1)=右边
又证:左边=2×5 n+2+3 n(32+1)-2 n(22+1)
=2×5 n+2+10×3 n-5×2 n
右边=10×5 n+1+10×3 n-10×2 n-1
=2×5 n+2+10×3 n-5×2 n
∴左边=右边
例2 己知:a+b+c=0 求证:a3+b3+c3=3abc
证明:∵a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-ac-bc)(见19例1)
∵:a+b+c=0
∴a3+b3+c3-3abc=0 即a3+b3+c3=3abc
又证:∵:a+b+c=0 ∴a=-(b+c)
两边立方 a3=-(b3+3b2c+3bc2+c3)
移项 a3+b3+c3=-3bc(b+c)=3abc
再证:由己知 a=-b-c 代入左边,得
(-b-c)3+ b3+c3=-(b3+3b2c+3bc2+c 3)+b3+c3
=-3bc(b+c)=-3bc(-a)=3abc
例3 己知a+ ,a≠b≠c 求证:a2b2c2=1
证明:由己知a-b= ∴bc=
b-c= ∴ca= 同理ab=
∴ab bc ca= =1 即a2b2c2=1
例4 己知:ax2+bx+c是一个完全平方式(a,b,c是常数)求证:b2-4ac=0
证明:设:ax2+bx+c=(mx+n)2 , m,n是常数
那么:ax2+bx+c=m2x2+2mnx+n2
根据恒等式的性质 得 ∴: b2-4ac=(2mn)2-4m2n2=0
丙练习20
1. 求证: ①(a+b+c)2+(a+b-c)2-(a-b-c)2-(a-b-c)2=8ab
②(x+y)4+x4+y4=2(x2+xy+y2)2 ③(x-2y)x3-(y-2x)y3=(x+y)(x-y)3
④3 n+2+5 n+2―3 n―5 n=24(5 n+3 n-1) ⑤a5n+a n+1=(a3 n-a2 n+1)(a2 n+a n+1)
2.己知:a2+b2=2ab 求证:a=b
3.己知:a+b+c=0
求证:①a3+a2c+b2c+b3=abc ②a4+b4+c4=2a2b2+2b2c2+2c2a2
4.己知:a2=a+1 求证:a5=5a+3
5.己知:x+y-z=0 求证: x3+8y3=z3-6xyz
6.己知:a2+b2+c2=ab+ac+bc 求证:a=b=c
7.己知:a∶b=b∶c 求证:(a+b+c)2+a2+b2+c2=2(a+b+c)(a+c)
8.己知:abc≠0,ab+bc=2ac 求证:
9.己知: 求证:x+y+z=0
10.求证:(2x-3)(2x+1)(x2-1)+1是一个完全平方式
11己知:ax3+bx2+cx+d能被x2+p整除 求证:ad=bc
1. 求证: ①(a+b+c)2+(a+b-c)2-(a-b-c)2-(a-b-c)2=8ab
②(x+y)4+x4+y4=2(x2+xy+y2)2 ③(x-2y)x3-(y-2x)y3=(x+y)(x-y)3
④3 n+2+5 n+2―3 n―5 n=24(5 n+3 n-1) ⑤a5n+a n+1=(a3 n-a2 n+1)(a2 n+a n+1)
2.己知:a2+b2=2ab 求证:a=b
3.己知:a+b+c=0
求证:①a3+a2c+b2c+b3=abc ②a4+b4+c4=2a2b2+2b2c2+2c2a2
4.己知:a2=a+1 求证:a5=5a+3
5.己知:x+y-z=0 求证: x3+8y3=z3-6xyz
6.己知:a2+b2+c2=ab+ac+bc 求证:a=b=c
7.己知:a∶b=b∶c 求证:(a+b+c)2+a2+b2+c2=2(a+b+c)(a+c)
8.己知:abc≠0,ab+bc=2ac 求证:
9.己知: 求证:x+y+z=0
10.求证:(2x-3)(2x+1)(x2-1)+1是一个完全平方式
11己知:ax3+bx2+cx+d能被x2+p整除 求证:ad=bc
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