Question

已知x^2-5-14=0,求代数式(x-1)(3x-1)-(x+3)^2+5的值

Answer 1

x^2-5x-14=0
即(x-7)(x+2)=0
则x=7或-2

(x-1)(3x-1)-(x+3)^2+5
=(7-1)(21-1)-(7+3)^2+5 或=(-2-1)(-6-1)-(-2+3)^2+5
=25

Answer 2

(x-1)(3x-1)-(x+3)^2+5
=3x²-x-3x+1-x²-6x-9+5
=2x²-10x+6
=2(x²-5x+3)
x^2-5-14=0
我在想你是不是输错了，应该是这样的吧
x^2+5-14=0
x²=9
x=±3
(x-1)(3x-1)-(x+3)^2+5
=2(x²-5x+3)
=2(9±15+3)
=24±30
=54或者-6

Answer 3

∵x²-5x-14=0
∴x²-5x=14
(x-1)(3x-1)-(x+3)²+5
=3x²-4x+1-(x²+6x+9)+5
=2x²-10x-3
=2(x²-5x)-3
=2×14-3
=25

Answer 4

先求1式，十字相乘法。解出x=7 或x=-2
2式化简：2x²-10x+6 带入即可。