已知M²+N-1=3,求1/3M²+1/3N-6。
解:由题意得:M²+N=4,所以:
1/3M²+1/3N-6
=1/3×(M²+N)-6
=1/3×4-6
=-14/3
m²-3m+1=0
(m-3/2)²=5/4
m1=3/2-√5/2
m2=3/2+√5/2
m²=3m-1
3m²-8m²+2
=-5m²+2
=-5X(3m-1)+2
=-15m+5+2
=-15m+7
m1=3/2-√5/2时,原式=-31/2+15√5/2
m2=3/2+√5/2时,原式=-31/2-15√5/2
解:-3m^2+11mn-14n^2-6mn+7m^2+10n^2-3=4m^2+5mn-4n^2-3
=4*13+5*(-6)-3=19
用夹逼定理
n²/(n²+n)<n[1/(n²+1)+1/(n²+2)+1/(n²+3)+……+1/(n²+n)]<n²/(n²+1)
lim n²/(n²+n)=limn²/(n²+1)=1
所以lim n[1/(n²+1)+1/(n²+2)+1/(n²+3)+……+1/(n²+n)]=1
3m^2+3m+2010=3(m^2+m-1)+3+2010=2013
m^2-3m+1=0
m-3+1/m=0
m+1/m=3
(m+1/m)^2=9
m^2+2+1/m^2=9
m^2+1/m^2=7
解
(1/2m-1/3n)²×(1/2m+1/3n)²
=[(1/2m-1/3n)(1/2m+1/3n)]²
=(1/4m²-1/9n²)²
=1/16m^4-1/18m²n²+1/81n^4
(x的平方加x减3/x平方加x减2)加1等于2x平方加4x加1/x平方加2x加1
(x²+x-3/x²+x-2)+1=2x²+4x+1/x²+2x+1
(x²+2x-2-3/x²)+1=2x²+6x+1+1/x²
x²+2x-2-3/x²+1-2x²-6x-1-1/x²=0
-x²-4x-2-4/x²=0
x^4+4x^3-2x^2-4=0
x1≈-4.48
x2=1.07
已知m、n是一元二次方程x²+x-1=0的根,则:
m²+m-1=0;n²+n-1=0;
且由一元二次方程根与系数的关系有:m+n=-1,mn=-1
那么,m²-mn+3m+n=m²+1+3m+n
=(m²+m-1)+2m+n+2
=0+m+(m+n)+2
=m-1+2
=m+1
——你确定题目中是加3m,而不是2m?!
解由丨m一2丨十(m十n)^2=0
知m-2=0且m+n=0
即m=2,n=-2
故原式=4m^2-6n^2-2mn-2m^2+3n
=2m^2-6n^2-2mn+3n
=2(2)^2-6(-2)^2-2*2*(-2)+3(-2)
=8-24+8-6
=-14