A math problem of a quadratic equation, a math problem of a quadratic equation

From Vedder's theorem, mn=7, m+n=-2008

Because m and n are two equations, m +2007m + 6 = -m - 1, n +2009n + 8 = n + 1

So the original formula is equal to -(m+1)*(n+1)=-(mn+m+n+1)=2000

Solution: Original formula = (-m-1) x (n + 1) = -mn-1 - (m + n).

According to the problem, the great theorem is obtained by mn=7, m+n=-2008

So the original formula = -7-1 + 2008 = 2000

m +2008m+7=0 m +2007m+6=-m-1n +2008n+7=0 n +2009n+8=n+1 The formula is =-(m+1)(n+1)=-(mn+m+n+1)mn=7 m+n=-2008, so the sought =-(-2008+7+1)=2000

m²+2007m+6）（n²+2009n+8）=（m²+2008m+7-m-1）（n²+2008n+7+n+1）=（-m-1）（n+1）=-（m+1）（n+1）=-（mn+m+n+1）

From the relationship between the root and the coefficient, we know that m+n=-2008,mn=7

So -(mn+m+n+1)=-(7-2008+1)=2008

It's a no-brainer. The proof is shown in the picture below.

Solution: Let the acceleration be a

Formula: vt 2-vo 2=2as

0-20^2=2a*25

a=-8m/s^2

The time to set the line for 15 meters is t

s=vot+1/2at^2

15=20t-1/2*8t^2

4t^2-20t+15=0

t=(20-4 root number 10) 8=(5-root number 10) 2, that is, the time is (5-root number 10) 2 seconds.

The original formula can be written as the square of [2(x-3)] minus the square of [5(x-2)], and then use the square difference formula, and the result is.

2((x-3)+5(x-2)】【2((x-3)-5(x-2)】=0, and then simplify it again.

The square of [2 (x 3)] - the square of [5 (x 2)] = 0

Use the square of a - the square of b = (a+b)(a-b).

Wouldn't it be nice to think of 4 (x 3) as a and 25 (x 2) as a b?

The square of [2(x-3)] x[5(x-2)]=o is sorted out to (7x-16)(4-3x)=0, and then it is directly obtained.