若X1,X2是方程X²-2X-5=0的两根 求(X1-X2)²

2024-12-25 17:15:04
推荐回答(3个)
回答1:

解:根据韦达定理:x1 + x2 = 2 , x1 * x2 = -5
所以(x1 - x2)²= (x1 + x2)² -4x1 * x2= 2² - 4×(-5)= 24

X1²+2X2-3 X1的三次方-9X1-10
= X1²-2X1+2X1+2X2-3 =X1(X1^2-2X1)+2X1^2-9X1-10
=X1²-2X1+2(X1+X2)-3 =5X1+2X1^2-9X1-10
=X1²-2X1+4-3 =2X1^2-4X1-10=0
=X1²-2X1-5+6
=6

回答2:

根据韦达定理
x1 + x2 = 2
x1 * x2 = -5
所以(x1 - x2)²
= (x1 + x2)² -4x1 * x2
= 2² - 4×(-5)
= 4 + 20
= 24

回答3:

先求出x1 x2 在带入(x1-x2)的平方 计算