15问答网 > 2x^2-7x 5=0 (x-2⼀X+2)-(10Y+5⼀X^2-4)=1 (2X-3)^2-2X+3=0 (X+1)(X-3)>=(X+1)^2

2x^2-7x 5=0 (x-2⼀X+2)-(10Y+5⼀X^2-4)=1 (2X-3)^2-2X+3=0 (X+1)(X-3)>=(X+1)^2

2025-05-17 18:13:41

推荐回答（2个）

回答1：

解：
1. 2x²-7x+5 = 0
2 -5
1 -1
(2x-5)(x-1) = 0 , x = 5/2 , 1

2. 怎么有个Y？

3. (2x-3)²-2x+3 = 0
(2x-3)²-(2x-3) = 0
(2x-3)(2x-4) = 0
x = 3/2 ,2

4.(x+1)(x+3) ≥ (x+1)²
(x+1)(x+3) - (x+1)² ≥ 0
(x+1)(x+3-x-1) ≥ 0
2(x+1) ≥ 0
x ≥ -1

回答2：

解：
1. 2x²-7x+5 = 0
2 -5
1 -1
(2x-5)(x-1) = 0 , x = 5/2 , 1

2. (2x-3)²-2x+3 = 0
(2x-3)²-(2x-3) = 0
(2x-3)(2x-4) = 0
x = 3/2 ,2

3.(x+1)(x+3) ≥ (x+1)²
(x+1)(x+3) - (x+1)² ≥ 0
(x+1)(x+3-x-1) ≥ 0
2(x+1) ≥ 0
x ≥ -1