用配方法解下列方程(1)2x 4x 3 0 2 2x 1 x 3 1 3 2x 2x

時間 2021-09-09 01:57:46

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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