n的平方加n减一等于0,n的3次方加2n的平方加2013等于?
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m²+n²-6m+10n+34=(m-3)²+(n+5)²=0所以m=3,n=-5
由题意得:m^2+n^2-6m+10n+34=0∴(m^2-6m+9)+(n^2+10n+25)=0∴(m-3)^2+(n+5)^2=0又(m-3)^2≥0,(n+5)^2≥0∴m=3,n=-5∴m+
m²-6m+9+n²-4n+4=0(m-3)²+(n-2)²=0∴m-3=0n-2=0∴m=3n=2再问:为什么(m-3)²+(n-2)²=
x^2-3x+1=0x^2+1=3x两边除Xx+1/x=3平方x^2+2+1/x^2=9x^2+1/x^2=7
m和n互为相反数,m+n=0,m=-n,n不等于0,则m也不等于0,mn²+m²n-2n/m=mn(n+m)-2n/(-n)=mn(0)+2=0+2=2
从m²-4m-1=0,n²-4n-1=0可知m和n分别为方程x²-4x-1=0的两个根所以m+n=4,mn=1m/n+n/m=(m²+n²)/mn=(
(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
根据极限定义:要证明极限为1,只需证明|(n^2-2)/(n^2+n+1)-1|
=[(3m+2n)+(m-n)][(3m+2n)-(m-n)]=(3m+2n+m-n)(3m+2n-m+n)=(4m+n)(2m+3n)
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
依题意得:原式化简为=m平方/(m平方-n平方)又因为3m=2n,所以代入化简后的式子,结果为-4/5
解m²=n+3n²=m+3两式相减得:m²-n²=n-m即(m-n)(m+n)=-(m-n)∵m≠n∴m+n=-1又m²-n=3,n²-m=
n^2+n-1=0可得:n^2+n=1n^3+2n^2+2009=n^3+n^2+n^2+2009=n(n^2+n)+n^2+2009=n+n^2+2009=1+2009=2010
㎡+n²+4m-6n+13=0㎡+4m+4+n²-6n+9=0(m+2)²+(n-3)²=0m=-2n=3(-2)³=-8再问:已知a减b等于5,ab
(m+1)方+(n-3)方=0m=-1n=3m+n=2
/>n²+n-800=0n²+n+1/4=800+1/4(n+1/2)²=3201/4n+1/2=±√(3201/4)n=-1/2±√3201/2n1=(-1+√3201
m^2+n^2-10m+6n+34=0(m^2-10m+25)+(n^2+6n+9)=0(m-5)^2+(n+3)^2=0m=5n=-3
n是偶数时,(-1)的N次方再减一等于0n是奇数时,(-1)的N次方再减一等于-2
再问:有没有解题思路哇==再答:仔细看一下解题思路就在过程当中的嘛!