m是方程x² x 1=0的根,则式子4m² 4m 2020
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/15 16:06:23
有题意有:x1x2=m,x1+x2=2根号2,2x1+x2=-3根号2+1解得x1=-5根号2+1,x2=7根号2-1,m=-71+12根号2;(根号下x1/x2)+(根号下x2/x1)无解,因为x1
x1,x2是x²+(2-M)x+(1+M)=0的两个根x1+x2=M-2x1x2=1+Mx1²+x2²>=2x1x2=2(1+M)当且仅当x1=x2时,有最小值.即根的判
所以x1+x2=13x1x2=m所以x2²-13x2+m=0x2²=13x2-m所以13x1+x2²+2=0所以13x1+13x2-m+2=013×13-m+2=0m=1
解题思路:由韦达定理可知x1+x2=6,又x2=2x1解方程组得x1=2,x2=4m=x1×x2=8解题过程:
1题x1+x2=-mx1x2=-1则x1^2+x2^2=(x1+x2)^2-2x1x2=m^2+2=17m=±√15判别式大于等于0m^2-4(m-1)>=0(m-2)^2>=0恒成立所以m=±√15
x^2-2mx+m+2=0△=4m^2-4(m+2)≥0m^2-m-2≥0(m-2)(m+1)≥0m≥2,m≤-1x1^2+x2^2=(x1+x2)^2-2x1x2=(2m)^2-2*2=4m^2-4
x1+x2=13x1x2=mx1+x2=(4x2-2)+x2=135x2=15x2=3x1=10m=x1x2=30
韦达定理x1+x2=-mx1x2=m-1所以x1²+x2²=(x1+x2)²-2x1x2=m²-2m+2=17m²-2m-15=(m-5)(m+3)=
把x=x1=m/3代入3m²/9-19m/3+m=0m²-22m=0m(m-22)=0m=0,m=22m=0则x1=0韦达定理x1+x2=19/3x2=19/3m=22则x1=22
x1,x2是方程x²-(2m-1)x+(m²+2m-4)=0的两个实数根所以x1+x2=2m-1,x1x2=m²+2m-4Δ=(2m-1)²-4(m²
由⊿=(-2m)²-4(1-m²)=8m²-4≥0,得m²≥1/2.又x1+x2=2mx1x2=1-m²则x1²+x2²=(x1+
1.两个不相等的实数根9-4m>0所以m
x1+x2=13x1-4x2+2=0所以x2=3x1=10所以m=x1x2=30
(1)由韦达定理:x1+x2=3,x1x2=mS=x1^2+x2^2=(x1+x2)^2-2x1x2=9-2m即为所求的解析式.方程有两个不同的实数根,所以判别式大于0判别式Δ=9-4m>0即m
x1+x2=-2,x1x2=mx1^2-x2^2=(x1+x2)(x1-x2)=2得x1-x2=-1两边平方,得(x1+x2)^2-4x1x2=14-4m=1,昨m=3/4
△=(2-m)^2-4(1+m)=4-4m+m^2-4-4m=m^2-8m≥0m(m-8)≥0m≤0或m≥8x1^2+x2^2=(x1+x2)^2-2x1x2=(m-2)^2-2(1+m)=m^2-4
1、x^2+4x-m^2+2m+3=(x+3-m)(x+1+m)=0,——》x1=m-3,x2=-m-1,——》-1
由韦达定理知:x1+x2=-1,则x1,x2不可能同为正数若x1
x1+x2=-2m+1x1*x2=m^2+1x1^2+x2^2=(x1+x2)^2-2x1x2=(-2m+1)^2-2(m^2+1)=4m^2-4m+1-2m^2-2=2m^2-4m-1=15得2m^