已知m的平方加3m减2等于零,求8 2m平方加6m等于多少
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根为二分之5加减根号5
m²+m-1=0则:m²+m=1又:m³+2m²+2012=(m³+m²)+m²+2012=m(m²+m)+m²
m^2-3m+1=0m-3+1/m=0m+1/m=3(m+1/m)^2=9m^2+2+1/m^2=9m^2+1/m^2=7
m^2+m-1=0,所以m^2+m=1m^3+2m^2+2012=m^3+m^2+m^2+2012=m(m^2+m)+m^2+2012=m+m^2+2012=2012+1=2013
(m^2+n^2)(m^2+n^2-9)-10=0,则(m^2+n^2)^2-9(m^2+n^2)-10=0,∴(m^2+n^2+1)(m^2+n^2-10)=0,m^2+n^2+1>0,∴m^2+n
m^2-5m-1=0,可得m^2=5m+12m^2-5m+1/m^2=10m+2-5m+1/(5m+1)=(25m^2+15m+3)/(5m+1)=(125m+25+15m+3)/(5m+1)=28
思路:把所求里逐渐凑成m^2+m+3,然后去掉即可m^5+3m^4+2m^3+2m^2-10m=m^5+m^4+3m^3+2m^4-m^3+2m^2-10m=m^3(m^2+m+3)+2m^4-m^3
设M(x1,y1),N(x2,y2),方程x²+y²-2x-4y+m=0与x+2y-4=0联立,(4-2y)²+y²-2(4-2y)-4y+m=05y²
根据题意知道:m^2-m-2014=0,则m^2=2014+m根据韦达定理有m+n=1m^2+n=2014+m+n=2015
m²-2m-1=0(1)n²-2n-1=0 (2)由(1)、(2)两式得:m、n是方程x²-2x-1=0的两个实数根.由韦达定理得: m+n=2m·n
Δ小于零再答:b平方减4ac
m是方程x²-5x-1=0的根所以:m²-5m-1=0m²=5m+1m=(5±√29)/2(2m²-5m+1)/m²=(10m+2-5m+1)/(5m
3m^2+3m+2010=3(m^2+m-1)+3+2010=2013
x^2-12x+m=0x^2-12x+36=36-m(x-6)^2=36-mx=6±根号(36-m)因为x2=6+根号(36-m)>6,并且一个根是另一个根的二倍,所以两根比都为正数,并且x2=2x1
该问题转化为:(m+a)(m-b)=0(n+c)(n-d)=0abcd则需满足:ab+cd=34b-a=6c-d=4则m+n=b+d=d-a=b-c=-a-cb-a-(c-d)=b-c+d-a=2(m
初四?m^4=(根号101)^2=112*根号10m^(-4)=1/112*根号10=11-2*根号10/((112*根号10)(11-2*根号10))=112*根号10/81m^4