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

2x^2-7x 5=0 (x-2/X+2)-(10Y+5/X^2-4)=1 (2X-3)^2-2X+3=0 (X+1)(X-3)>=(X+1)^2
2x^2-7x 5=0
(x-2/X+2)-(10Y+5/X^2-4)=1
(2X-3)^2-2X+3=0
(X+1)(X-3)>=(X+1)^2
用十字相乘法.
maltc 1年前 已收到2个回答 举报
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cwf372002 花朵

共回答了20个问题采纳率：90% 举报

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

1年前

7

jinjin1234 幼苗

共回答了1个问题 举报


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 =...

1年前

0
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