1樓:新野旁觀者
(1)2x²+4x-3=0
x ²+2x+1=5/2
(x+1)²=5/2
x+1=±1/2√10
x=-1±1/2√10
﹙2﹚﹙2x+1﹚﹙x-3﹚=1
2x²-5x=4
x²-5/2x=2
x-2/5x+25/16=57/16
(x-5/4)²=57/16
x-5/4=±1/4√57
x=5/4±1/4√57
﹙3﹚2x²+√2x-3=0
x²+1/2√2x+(1/4√2)²=3+1/8(x+1/4√2) ²=25/8
x+1/4√2=±5/4√2
x=-1/4√2±5/4√2
﹙4﹚½x²+x-3=0
x²+2x+1=7
(x+1)²=7
x+1=±√7
x=-1±√7
2樓:世翠巧
解:2x²+4x-3=0
2x²+4x=3 方程兩邊同時乘1/2x²+2x=3/2
x²+2x+1=1+3/2
(x+1)²=5/2
x+1=±(√10)/2
x=(-2±√10)/2
x1=(-2+√10)/2
x2=(-2-√10)/2
(2x+1)(x-3)=1
2x²-5x-3=1
2x²-5x=4 方程兩邊同時乘1/2x²-5x/2=2
x²-5x/2+(5/4)²=2+(5/4)²(x-5/4)²=57/16
x-5/4=±(√57)/4
x=(5±√57)/4
x1=(5+√57)/4
x2=(5-√57)/4
2x²+√2x-3=0
2x²+√2x=3 方程兩邊同時乘1/2x²+√2x/2=3/2
x²+√2x/2+(√2/4)²=3/2+(√2/4)²(x+√2/4)²=13/8
x+√2/4=±(√26)/4
x=(-√2±√26)/4
x1=(-√2+√26)/4
x2=(-√2-√26)/4
½x²+x-3=0 方程兩邊同時乘2x²+2x-6=0
x²+2x=6
x²+2x+1=6+1
(x+1)²=7
x+1=±√7
x=-1±√7
x1=-1+√7
x2=-1-√7
用一元二次方程解下列方程(1)y=5(x-1)² (2)y=2x²-4x-1(3)y=3x²-6x+2(4)y=(x+1)(x-2)
3樓:匿名使用者
設y為已知,可得
(1) y=5(x-1)²
(x-1)²=y/5
x-1=±√(y/5)
x=±√(y/5)+1
(2) y=2x²-4x-1
y=2(x²-2x-1/2)=2(x²-2x+1-3/2)=2(x-1)²-3
2(x-1)²=y+3
(x-1)²=(y+3)/2
x-1=±√(y+3)/2
x=±[√(y+3)/2]+1
(3) y=3x²-6x+2 y=3(x²-2x+2/3)=3(x²-2x+1-1/3)=3(x-1)²-1
3(x-1)²=y+1
(x-1)²=(y+1)/3
x-1=±√(y+1)/3
x=±[√(y+1)/3]+1
(4) y=(x+1)(x-2)=x²-x-2=x²-x+1/4-9/4=(x-1/2)²-9/4
(x-1/2)²=y+9/4
x-1/2=±√(y+9/4)
x=±[√(y+9/4)]+1/2
(5) 無等式,無解
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