一元二次方程计算题1.(x-5)^2=2(x-5)2.x^2-(√2)*x-1/4=03.x^2+(22√5)*x+10=04.已知a,b是方程x^2+x-1=0的两根,求a^2+2a+1/b的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 19:09:15
![一元二次方程计算题1.(x-5)^2=2(x-5)2.x^2-(√2)*x-1/4=03.x^2+(22√5)*x+10=04.已知a,b是方程x^2+x-1=0的两根,求a^2+2a+1/b的值](/uploads/image/z/1345479-15-9.jpg?t=%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8B%E8%AE%A1%E7%AE%97%E9%A2%981.%28x-5%29%5E2%3D2%28x-5%292.x%5E2-%28%E2%88%9A2%29%2Ax-1%2F4%3D03.x%5E2%2B%2822%E2%88%9A5%29%2Ax%2B10%3D04.%E5%B7%B2%E7%9F%A5a%2Cb%E6%98%AF%E6%96%B9%E7%A8%8Bx%5E2%2Bx-1%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%2C%E6%B1%82a%5E2%2B2a%2B1%2Fb%E7%9A%84%E5%80%BC)
一元二次方程计算题1.(x-5)^2=2(x-5)2.x^2-(√2)*x-1/4=03.x^2+(22√5)*x+10=04.已知a,b是方程x^2+x-1=0的两根,求a^2+2a+1/b的值
一元二次方程计算题
1.(x-5)^2=2(x-5)
2.x^2-(√2)*x-1/4=0
3.x^2+(22√5)*x+10=0
4.已知a,b是方程x^2+x-1=0的两根,求a^2+2a+1/b的值
一元二次方程计算题1.(x-5)^2=2(x-5)2.x^2-(√2)*x-1/4=03.x^2+(22√5)*x+10=04.已知a,b是方程x^2+x-1=0的两根,求a^2+2a+1/b的值
1.(x-5)^2=2(x-5)
(x-5)^2-2(x-5)=0
(x-5)(x-5-2)=0
(x-5)(x-7)=0
x=5,x=7
2.x^2-(√2)*x-1/4=0
x^2-√2x+1/2=1/4+1/2=3/4
(x-√2/2)^2=(±√3/2)^2
x-√2/2=±√3/2
x=(√2+√3)/2,x=(√2-√3)/2
3.x^2+(22√5)*x+10=0
x^2+22√5x+(11√5)^2=-10+(11√5)^2=595
(x+11√5)^2=(±√595)^2
x+11√5=±√595
x=-11√5+√595,x=-11√5-√595
4.已知a,b是方程x^2+x-1=0的两根,求a^2+2a+1/b的值
a是方程的根
所以a^2+a-1=0
a^2+a=1
a^2+2a=a^2+a+a=a+1
所以a^2+2a+1/b=a+1+1/b=(ab+b+1)/b
由韦达定理,ab=-1
所以a^2+2a+1/b=(ab+b+1)/b=(-1+b+1)/b=b/b=1