Math Questions! Ask God for advice on math problems!

1 solution: Since m m 1 0, m = 1 m, m = m m, substituted m 2m 2011 = -m m + 2m + 2011 = 2012

2 solution: x m x 15 (x 3)(x n), 3n=-15, n=-5;3+n=m, we get m=-2, m to the nth power -1 32

Question 1: m 3 + 2m 2 + 2011 = m 3 + m 2-m + m + m 2 + 2011 = m (m 2 + m -1) + m 2 + m -1 + 2012

Substituting m 2 + m - 1 = 0 into it, the final result 2012 Question 2 is obtained: because: x 2 + mx - 15 = (x + 3) (x + n) = x 2 + (n + 3) + 3n

So we get the binary system of equations (1)--m=n+3

2)--3n=-15

The solution shows that m=2 n=-5

So m to the nth power = 1 32

1) m m 1 0 to get m m = 1

m³＋2m²＋2011

m^3-1)+2m^2+2012

m-1)(m^2+m+1)+2m^2+20122（m-1）+2m^2+2012

2(m^2+m)+2010

2) m = n + 3 (x 3) (x n = x + (n + 3) x + 3n from x mx 15 (x 3) x + 3 (x 3)The solution gives m = -2, n = -5 so m to the n power = (-2) (5) = -1 32

Suppose the abscissa is x1.

The y of the two straight lines is y1=1 2x1+3, y2=-1 6x1+5, and y1-1=y2 after moving y1 downwards by one unit.

So get the equation 1 2x+3-1=-1 6x+5, and you can calculate x=9 2

Because x is required, and because the first one is high, the two formulas are subtracted by 1 = 1 2x+3-(-1 6x+5), and the abscissa of the solution a is the first netizen to answer correctly.

1=1 2x+3-(-1 6x+5), the abscissa of a is .

9 2, this question needs to draw an extension line!

5.Solution:

Let x 4 6x 13x kx 4 (x px q) x 4 2px (p 2q) x 2pqx q make both sides of the equation equal if and only if the coefficients are equal.

Get the system of equations:

2p＝－6p²＋2q＝13

2pq＝kq²＝4

Solution: p 3, q 2, k 12

Answer: The value of the constant k is 12.

6.Solution: Let x x 8x 12 (x p)(x q) x (p 2q)x (2pq q )x pq make both sides of the equation identant, if and only if the coefficients are equal.

Get the system of equations:

p＋2q＝－1

2pq＋q²＝－8

pq²＝12

Solution: p 3, q 2

x³－x²－8x＋12＝(x＋3)(x－2)²∴x1＝－3，x2＝x3＝2

5.With the method of pending coefficients.

Reams x 4-6x 3+13x 2+kx+4

x^2+ax+b)^2

x 4+2ax 3+(a 2+2b) x 2+abx+b 2 to get 2a=-6

a^2+2b=13

k=abb^2=4

The solution yields a=-3, b=2, k=-6

k=-66.With the method of pending coefficients.

x^3-x^2-8x+12=0

x-a)^2(x-b)=0

x^2-2ax+a^2)(x-b)=0

x 3-(2a+b) x 2+(a 2+2ab)x-a 2b=0 comparison.

2a+b=1

a(a+2b)=-8

a^2b=-12

The solution is a=2 b=-3

The solution of the equation is x1 = x2 = 2 and x3 = -3

Both of these questions are tested on the pending coefficient method.

f(x)=(1+ax^2)/(x+b)

According to the nature of the odd function there is.

f(x)=-f(-x)

Get: (1+ax 2) (x+b)=-1+ax 2) (x+b) i.e.: (1+ax 2) (x+b)=(1+ax 2) (x-b) So b=0 and then substitute (1,3) points into have.

1+a1^2)/1=3

Solution; a=2

So the function is: f(x)=(1+2x 2) x1 x+2x

When x>0, f(x)=1 x+2x>=2 2 is based on the property of the odd function, with respect to the principle of origin symmetry.

When x<0, f(x)=1 x+2x<=-2 2

(1)sn/tn=(7n+2)/(n+3)sn/(7n+2)=tn/(n+3) =ksn =(7n+2)k , tn= (n+3)ka8/b8

s8-s7)/(t8-t7)

58k-51k)/(11k-10k)

let an = sum of the lengths of the sides of the nth regular hexagon.

a1=6ra2/6)^2 = (r/6)^2 + r/6)^2 - 2(r/6)(r/6)cos(2π/3)

3r^2/36

a2 = √3r

an/6)^2 = (a(n-1)/6)^2 + a(n-1)/6)^2 - 2(a(n-1)/6)(a(n-1)/6)cos(2π/3)

3a(n-1)^2/36

an= √3a(n-1)

an/a(n-1) = √3

an/a1= (√3)^(n-1)

an= 6r (√3)^(n-1)

sn = a1+a2+..an

6r[(√3)^n -1] /(√3-1 )

M must be present such that sn=m(7n+2), tn=m(n+3)a8=s8-s7=m(58-51)=7m

b8=t8-t7=m(11-10)=m

a8/b8=7

The side length and the latter one are 3 2 of the previous one

an=6r*（√3/2）^n

sn=a1(1-q^n)/(1-q)

3 3r(1-( 3 2) n) (1- 3 2) When n tends to infinity, ( 3 2) n = 0

The sum of the side lengths of all these regular hexagons.

s=3√3r/(1-√3/2)

12√3+18)r