Question

几道数学题：1、已知x+1/x=5,求x-1/x. 2、已知ab/(a+b)=1/2,bc/(b+c)=1/3,ac/(a+c)=1/4,求a、b、c的值。

3、因式分解：（a^2-b^2-c^2)^2-4b^2c^2 ＜答案：（a+b-c)(a-b+c)(a+b+c)(a-b-c) ＞
4、已知a/b=c/d=e/f,求（a^n+c^n+e^n)/(b^n+d^n+f^n)-(a+c+e)^n/(b+d+f)^n的值。
请各位大哥大姐帮忙一下，最主要的是过程啊！

Answer 1

1、已知x+1/x=5,求x-1/x
解：(x+1/x)²=x²+2+1/x²=25，故x²+1/x²=23；
(x-1/x)²=x²-2+1/x²=23-2=21，∴x-1/x=±√21.
2、已知ab/(a+b)=1/2,bc/(b+c)=1/3,ac/(a+c)=1/4,求a、b、c的值。
解：2ab=a+b，故1/a+1/b=2.........(1)
3bc=b+c， 故1/b+1/c=3..........(2)
4ac=a+c， 故1/a+1/c=4...........(3)
(1)+(2)+(3)得2(1/a+1/b+1/c)=9，故1/a+1/b+1/c=9/2..........(4)
将(1)代入(4)式得1/c=9/2-2=5/2，故c=2/5；
将(2)代入(4)式得1/a=9/2-3=3/2，故a=2/3；
将(3)代入(4)式得1/b=9/2-4=1/2，故b=2.
3、因式分解：（a²-b²-c²)²-4b²c²
解：原式=(a²-b²-c²)²-(2bc)²=(a²-b²-c²+2bc)(a²-b²-c²-2bc)=[a²-(b²-2bc+c²)][a²-(b²+2bc+c²)]
=[a²-(b-c)²][a²-(b+c)²]=(a+b-c)(a-b+c)(a+b+c)(a-b-c)
4、已知a/b=c/d=e/f,求（a^n+c^n+e^n)/(b^n+d^n+f^n)-(a+c+e)^n/(b+d+f)^n的值。
解：设a/b=c/d=e/f=m，那么(a/b)ⁿ=(c/d)ⁿ=(e/f)ⁿ=mⁿ
由等比定理得(aⁿ+cⁿ+eⁿ)/(bⁿ+dⁿ+fⁿ)=aⁿ/bⁿ=(a/b)ⁿ=mⁿ
(a+c+e)/(b+d+f)=a/b=m，故(a+c+e)ⁿ/(b+d+f)ⁿ=mⁿ
∴ (aⁿ+cⁿ+eⁿ)/(bⁿ+dⁿ+fⁿ)-(a+c+e)ⁿ/(b+d+f)ⁿ=mⁿ-mⁿ=0

Answer 2

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Answer 3

楼上的第一题马虎了,应该是:
1、已知x+1/x=5,求x-1/x
解：(x+1/x)²=x²+2X+1/x²=25，
24x^2-2x-1=0
4 -1
6 1
十字相乘法得(4X-1)(6X+1)=0
解得X=1\4或X= -1\6
再将X值分别带入X-1\X即可得两个结果

Answer 4

因式分解的把后面的写成2bc*2bc,两个平方的差求因式分解，不难了吧。（4）设a/b=c/d=e/f=k;然后a=b*k,c=d*k,e=f*k,要是到这里你还不知道怎么做，你可以安心地去了