1、已知x+1/x=5,求x-1/x
(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.
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(a²-b²-c²)²-4b²c²
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.
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