Question

1.若实数X Y Z满足X+1/Y=4,Y+1/Z=1,Z+1/X=7/3,求XYZ乘积~

Answer 1

(x+1/y)(y+1/z)(z+1/x)=28/3

将其展开为

xyz+x+y+z+1/x+1/y+1/z+1/xyz=28/3①

将已知的三个因式相加得到

xyz+x+y+z+1/x+1/y+1/z=22/3②

①-②=2

即xyz+1/xyz=2

假设xyz=a

则a+1/a=2

得 a2-2a+1=0(a2表示a的平方)

得 a=1

即 x*y*z=1

A/x-1+B/x+2=(Ax+2A+Bx-B)/(x-1)(x+2)

=[(A+B)x+2A-B]/(X-1)(X+2)

(2X-3)/(X-1)(X+2)=[(A+B)X+2A-B]/(X-1)(X+2)

{A+B=2

{2A-B=-3

所以A=-1/3 B=7/3