(6-k)(9-k)x^2-(117-15k)x+54=0求k=?

∵方程(6-k)(9-k)x^2-(117-15k)x+54=0的两根均是整数
⊿=(117-15k)^2-216(6-k)(9-k) =117^2-2•117•15k+225k^2-216(54-15k+k^2)
=13689-3510k+225k^2-11664+3240k-216k^2
=9(k-15)^2>0
X1=[(117-15k)-3(k-15)]/[2(6-k)(9-k)] =18(9-k)/[2(6-k)(9-k)] =9/(6-k)
X2=[(117-15k)+3(k-15)]/[2(6-k)(9-k)] =12(6-k)/[2(6-k)(9-k)] =6/(9-k)]
若x1为整数，6-k可为-1,-3,-9,1,3,9, 则k=7,9,15,5,3,-3
若x2为整数，9-k可为-1,-2,-3,-6,1,2,3,6则k=10,11,12,15,8,7,6,3
综上：使二根同时为整数的k值为3，15

(6-k)(9-k)x^2-(117-15k)x+54=0求k=?

△ =(117-15k)²-4*(6-k)(9-k)*54=0
=117²-2*117*15k+15²k²-4(54-6k-9k+k²)*54
=137709-36270k+225k²-216(54-15k+k²)
实在很大的数据 你自己算啊
解出两个结果 然后判断有无意义就可以取舍了