m,n is the root of x 2 2011 x 2012 0, find the value of m 2 2010m 2013 n 2 2010n 2011

x^2+2011x-2012=(x-1)(x+2012) =0m,n) = (1, -2012)

m^2+2010m-2013)(n^2+2010n+2011) = (m^2 +2011m -2012 -m -1) (n^2 +2011n -2012 -n +4023)

m+1)(n-4023)

Apparently it has two possible values.

When m=1, n=-2012 brings in to get 2 (-2012-4023) and when m=-2012, n=1, brings in to get -2011*4022).

Solution: Because m,n is the root of x 2 + 2011x-2012 = 0, then m 2 + 2011 m - 2012 = 0, n 2 + 2011n - 2012 = 0

So m 2 + 2011 m = 2012 , n 2 + 2011n = 2012

So (m 2 + 2010 m - 2013) (n 2 + 2010n + 2011).

by x 2 + 2011 x - 2012 = 0

x-1)(x+2012)=0

Get x1=1 , x2=--2012

m^2+2010m-2013)(n^2+2010n+2011)(m^2+2010m-2011-2)(n^2+2010n-2011+4022)

m-1)(m+2011)-2][(n-1)(n+2011)+4022]

Substitute m=1 n=--2012 into the equation (or m=--2012 n=1 will result the same).

Answer: 1 because: m,n are both the roots of the equation +2010x-2012=0, so.

m}^+2012m-2012=0

+2010n-2012=0

So: +2012m-2011) ( 2010n-2011}( 2012m-2012+1) ( 2010n-2012+1}

The answer is -1, and the answer is as follows:

Because m and n are both the roots of the equation x 2 + 2007x-2009 = 0.

Therefore, m 2 + 2007m - 2009 = 0, that is, m 2 + 2007m = 2009

n 2 + 2007n - 2009 = 0, that is, n 2 + 2007n = 2009 so. m^2+2007m-2008)(n^2+2007n-2010)=(2009-2008)*(2009-2010)=1*(-1)=-1

From m and n are the two roots of the equation x2-2010x+2011=0, according to the definition of the root of the equation and the relationship between the root and the coefficient of the unary quadratic equation, we can get n2-2010n+2011=0, m2-2010m+2011=0, m+n=2010, mn=2011, and (n2-2011n+2012) (m2-2011m+2012)=(n2-2010n+2011+1-n) (m2-2010m+2011+1-m), you can find the answer

Vedic theorem. m+n=2011

m-n=-2011

mn=-2012

m*(Huai negotiation-n) = -2012

So -m and lead to -n are the 2 roots of the equation: x +2011x-2012=0.

Then the waiter is substituted.

m)²+2011m-2012=0

m²+2011m=2012

Similarly. n²+2011n=2012

Algebraic formula = (2012-2011) (2012-2013) = 1 (-1) = -1

Solution: m, n are the two brigade limbs of the equation Qi split x 2-2004x-2001=0.

then m 2-2004m-2001=0 and high town closed n 2-2004n-2001=0

m+n=2004,nm=-2001

m^2-2005m+3)(n^2-2005n+3)(m^2-2004m-2001-m+2004)(n^2-2004n-2001-n+2004)

2004-n)(2004-m)

mn-2004*(m+n)+2004^2

It is obtained from the root and coefficient correlation system (Wei Tuan's theorem).

m+n=2004，m*n=-2001

Substituting m into the equation x -2004x-2001 = 0 yields.

m²-2004m-2001=0

m²-2004m=2001

m²-2005m=2001-m

m reed or ant - 2005m+3=2001-m+3=2004-m, the same is true for n-2005n+3=2004-n

So. m²-2005m+3)(n²-2005n+3)(2004-m)(2004-n)

2004²-2004m-2004n+m*n2004²-2004(m+n)+m*n

Substituting x=m and x=n into the original equation respectively yields:

m^2+2010m+2011=0

n^2+2010n+2011=0

m^2+1010m+1010=-1

n^2+1010n+1012=1

m^+2010m+2010)(n^+2010n+2012)

m 2 + 2009m-2010 = m 2 + 2009m - 2009-1 = 0 is m 2 + 2009m-2009 = 1

n 2 + 2009n + 2010 = n 2 + 2009n - 2010 + 4020 = 0 is n 2 + 2009n - 2010 = -4020

So their product is -4020

If you understand the meaning of the root of the equation, you can understand the equation! I don't understand, you can ask me again.