初三解方程
3个回答
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1. 2χ²-10χ=3
χ²-5χ-3/2 = (x - 5/2)² -25/4 -3/2 = 0
(x - 5/2)² = 31/4
x - 5/2 = (根号31)/2 或 x - 5/2 = -(根号31)/2
x = [5+(根号31)]/2 或 x = [5-(根号31)]/2
2. 2(χ-1)² = 3
(χ-1)² = 3/2
x - 1 = (根号 6)/2 或 x - 1 = -(根号 6)/2
x = [2+(根号 6)]/2 或 x = [2-(根号 6)]/2
3. (χ+3)²=(1-2χ)²
x+3 = 1-2x 或 x+3 = -(1-2x)
3x = -2 或 x = 4
x = -2/3 或 x = 4
χ²-5χ-3/2 = (x - 5/2)² -25/4 -3/2 = 0
(x - 5/2)² = 31/4
x - 5/2 = (根号31)/2 或 x - 5/2 = -(根号31)/2
x = [5+(根号31)]/2 或 x = [5-(根号31)]/2
2. 2(χ-1)² = 3
(χ-1)² = 3/2
x - 1 = (根号 6)/2 或 x - 1 = -(根号 6)/2
x = [2+(根号 6)]/2 或 x = [2-(根号 6)]/2
3. (χ+3)²=(1-2χ)²
x+3 = 1-2x 或 x+3 = -(1-2x)
3x = -2 或 x = 4
x = -2/3 或 x = 4
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(1) 2χ²-10χ=3
2χ²-10χ-3=0
χ = (5±√31)/2
(2) 2(χ-1)²=3
2χ²-4χ+2-3=0
2χ²-4χ-1=0
χ = (2±√6)/2
(3) (χ+3)²=(1-2χ)²
χ²+6χ+9 = 1-4χ+4χ²
3χ²-10χ-8=0
(3χ+2)(χ-4)=0
χ=-2/3 或 χ=4
2χ²-10χ-3=0
χ = (5±√31)/2
(2) 2(χ-1)²=3
2χ²-4χ+2-3=0
2χ²-4χ-1=0
χ = (2±√6)/2
(3) (χ+3)²=(1-2χ)²
χ²+6χ+9 = 1-4χ+4χ²
3χ²-10χ-8=0
(3χ+2)(χ-4)=0
χ=-2/3 或 χ=4
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(1) 2χ²-10χ=3
2χ²-10χ-3=0
χ = (5±√31)/2
(2) 2(χ-1)²=3
2χ²-4χ+2-3=0
2χ²-4χ-1=0
χ = (2±√6)/2
(3) (χ+3)²=(1-2χ)²
χ²+6χ+9 = 1-4χ+4χ²
3χ²-10χ-8=0
(3χ+2)(χ-4)=0
χ=-2/3 或 χ=4
2χ²-10χ-3=0
χ = (5±√31)/2
(2) 2(χ-1)²=3
2χ²-4χ+2-3=0
2χ²-4χ-1=0
χ = (2±√6)/2
(3) (χ+3)²=(1-2χ)²
χ²+6χ+9 = 1-4χ+4χ²
3χ²-10χ-8=0
(3χ+2)(χ-4)=0
χ=-2/3 或 χ=4
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