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)^22x^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.

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 14:35:03
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)^22x^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.

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)^22x^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.
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.

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)^22x^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.

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. 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 =...

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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

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