1.已知x1、x2为方程x^2+5x-1=0的两个实数根 则x1^2-5x2+17=?2.若实数m满足m^2-根号10m+1=0,求m^4+m^-4的值3.因式分解:x^2-5x-6
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 14:20:59
![1.已知x1、x2为方程x^2+5x-1=0的两个实数根 则x1^2-5x2+17=?2.若实数m满足m^2-根号10m+1=0,求m^4+m^-4的值3.因式分解:x^2-5x-6](/uploads/image/z/465758-62-8.jpg?t=1.%E5%B7%B2%E7%9F%A5x1%E3%80%81x2%E4%B8%BA%E6%96%B9%E7%A8%8Bx%5E2%2B5x-1%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%95%B0%E6%A0%B9+%E5%88%99x1%5E2-5x2%2B17%3D%3F2.%E8%8B%A5%E5%AE%9E%E6%95%B0m%E6%BB%A1%E8%B6%B3m%5E2-%E6%A0%B9%E5%8F%B710m%2B1%3D0%2C%E6%B1%82m%5E4%2Bm%5E-4%E7%9A%84%E5%80%BC3.%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%EF%BC%9Ax%5E2-5x-6)
1.已知x1、x2为方程x^2+5x-1=0的两个实数根 则x1^2-5x2+17=?2.若实数m满足m^2-根号10m+1=0,求m^4+m^-4的值3.因式分解:x^2-5x-6
1.已知x1、x2为方程x^2+5x-1=0的两个实数根 则x1^2-5x2+17=?
2.若实数m满足m^2-根号10m+1=0,求m^4+m^-4的值
3.因式分解:x^2-5x-6
1.已知x1、x2为方程x^2+5x-1=0的两个实数根 则x1^2-5x2+17=?2.若实数m满足m^2-根号10m+1=0,求m^4+m^-4的值3.因式分解:x^2-5x-6
1)因为x1,x2为方程x^2+5x-1=0的实数根,
所以x1^2+5x1-1=0,x1+x2=-5
所以x1^2=1-5x1,代人到x1^2-5x2+17得,
x1^2-5x2+17
=(1-5x1)-5x2+17
=1-5x2-5x2+17
=18-5(x1+x2)
=18-5*(-5)
=18+25
=43
2)因为m=0时,m^2-根号10m+1≠0,
所以方程两边除以m,得,
m-√10+1/m=0,
所以m+1/m=√10,平方,得
m^2+2+1/m^2=10,
所以m^2+1/m^2=8,再平方,
m^4+2+1/m^4=64,
所以m^4+m^-4=62,
3)x^2-5x-6=(x-6)(x+1)
1.
x1,x2是方程的根,则均满足方程。
x1^2+5x1-1=0
x1^2=1-5x1
又由韦达定理,得x1+x2=-5
x1^2-5x2+17
=1-5x1-5x2+17
=18-5(x1+x2)
=18-5(-5)
=43
2.
m^2-√10m+1=0
由韦达定理,得m1m2=1,即...
全部展开
1.
x1,x2是方程的根,则均满足方程。
x1^2+5x1-1=0
x1^2=1-5x1
又由韦达定理,得x1+x2=-5
x1^2-5x2+17
=1-5x1-5x2+17
=18-5(x1+x2)
=18-5(-5)
=43
2.
m^2-√10m+1=0
由韦达定理,得m1m2=1,即m1,m2互为倒数.
m1+m2=-√10
m^4+m^(-4)
=m1^4+m2^4
=(m1^2+m2^2)^2-2
=[(m1+m2)^2-2]^2-2
=[(√10)^2-2]^2-2
=(10-2)^2-2
=64-2
=62
3.
x^2-5x-6=(x-6)(x+1)
收起
1、18+5√29或者18-5√29
2、62
3、(x+1)(x-6)