Math problem solving, master in, there is a reward

1、/3x-2/-/x+1/=x+2

x<-1, 2-3x+x+1=x+2==> 3x=1 ==> x=1 3 (rounded).

1 x 2 3, 2-3x-x-1=x+2 ==> 5x=-1 ==>x=-1 5

x>2 3, 3x-2-x-1=x+2 ==> x=5

The solution of the original equation is x=-1 5 or x=5

2、/3x-/2x-1//=2

3x-|2x-1|=2 or 3x-|2x-1|=-2

x 1 2, 3x-2x+1=2 or 3x-2x+1=-2

x=1 or x=-3 (rounded).

x 1 2, 3x-2x+1=2 or 3x-2x+1=-2

x=3 5 (rounded) or x=-1 5

So the original equation is solved as x=1, or x=-1 5

3. If x=-1 is always the solution of equation 1-(mx+b) 3=(2x-am) 4 about x, try to find the values of a and b.

Substituting x=-1 into the original equation yields 1-(-m+b) 3=(-2-am) 4 constantly.

12+4m-4b=-6-3am.

For any m, the coefficients of the same term m on both sides of the equation are equal.

3a=4 and 12-4b=-6

a=-4/3,b=9/2

1. Order 3x-2 0 x 2 3Then x+1>03x-2-x-1=x+2 x=1,x 2 3 is the original solution.

Let x+1 0, x -1 3x-2<0

2-3x+x+1=x+2 x=1 3 x does not belong to x-1 is not the original solution.

Then when -12-3x-x-1=x+2 x=-1 5 is the solution of the original equation.

3. The original equation can be reduced to 6x+6b+(4x-3a)m=12 then x=1, 4x-3a=0, 6x+6b=12 then a=4 3, b=1

There is no solution, odd plus odd plus odd or odd.

This question is relatively simple.

Refer to the above ** Hope it can help you.

This problem is relatively simple, use parallel lines and parallelograms to determine AEDB as a parallelogram, AB is equal to ED, use 60 degree angles and two perpendicular to determine the similarity of two triangles, and then use the trigonometric ratio to find the length of the hypotenuse of the large triangle, the hypotenuse length is equal to the length of the opposite side of the two parallelograms, because the opposite sides of the AEDB parallelogram are equal. In the end, the eternal resemblance is longer than the ag.

Solution: (1) any x, f(-1+x)=a(-1+x) 3+b(-1+x) 2+c(-1+x)+d, f(-1-x)=a(-1-x) 3+b(-1-x) 2+c(-1-x)+d, the two formulas are added and sorted out 0Certification.

2) b = 0f (x) = ax 3 + cx + d, derivative 3ax 2 + c, derivative 0x 2 = -c (3a).

The value point can be present at x=0,x=1,x2=-c (3a).Let a=kc, d=mc, then f(0)=d=mc;

f(1)=a+c+d=(k+m+1)c,f(x=root(-1 (3k))).

The domain of f(x) is x>0, and if there is any 01 in the domain of definition, then ln(x2 x1) >0, and a>0, then f(x2)-(x1)>0

Therefore, the function increases monotonically.

When the extreme point is the minimum:

f'(x)=1/x+a/x^2， f''(x)=-1/x^2-2a/x^3

f'(x)=0, 1 x+a x 2=0, x=-a

f(-a)=ln(-a)-a/(-a)=ln(-a)+1

If ln(-a)+1=2, then a=-e, where x=e is in the interval [1,e], f''(e) = 1 e 2>0, i.e., there is a minima.

When the boundary value x=1 is the minimum value of the function:

f(1)=ln1-a=2, then a=-2

In this case, the extreme point f(-a)=f(2)=ln2+2 2=ln2+1<2, that is, it is smaller than the boundary value, so f(1) is not the minimum value of the function.

Hence a=-e