2 math problems, urgent, find two math problems urgent .

1.The answers of everyone upstairs are already OK, if it is a fill-in-the-blank question, you can use a special value x = 1 or 0 to substitute the solution, if it is a big problem, you must have a rigorous step:

m/（x+3）- n/（x-3）

m(x-3)- n(x+3)] (x 2 - 9)[(m-n)x - 3(m+n)] (x 2 - 9) To make the left side constant equal to the right side for any x, then (m-n)x - 3(m+n) is always equal to 8x, and the coefficients of each term are required to correspond to equal in turn, i.e., (m-n) = 8 and 3(m+n) = 0, and the solution is m = 4, n = -4, that is, mn = -16

2.Because x z, y z, so x + y > z + z, i.e. x + y > 2zTherefore, choose c (the same unequal sign can be added, if it is not in the same direction, it will not work, if it is x z, y z, you can't judge the relationship between x+y and z).

m (x+3) -n (x-3)=(m-n)x-3(m+n)=8x to be constant, then m-n=8 3(m+n)=0, solve the sub-binary equation to obtain m=4, n=-4 mn=-16

Add the two formulas to get x+y 2z and choose c

1.Because for any x, substituting x=1 and x=2 into the solution gives m=, so mn=-16

2.Choose C. When z>0, a,b are incorrect, and when x<<0, d is incorrect.

1.Because x≠ 3

The original is m(x-3)-n(x+3)=8x

mx-3m-nx-3n=8x

m-n-8)x=3m+3n

Because for any x(x≠ 3) is constant, we might as well let x=1, then we have m-n-8=3m+3n, 2m+4n=-8

Let x=-1, then there is m-n-8=-3m-3n, 4m+2n=8m=4, n=-4

mn=-16

Substituting x=0 gives m=-n

Then substituting m=-n, x=1 gives m=4, mn=-16, y z, so x-z 0, y-z 0, so (x-z)+(y-z) 0d (x-z)+y, y size is unknown.

From the image of the function, it can be seen that when x = 2 or -2, the function is maximum, and y = 4-2t-1 = 3-2t, because it is an open interval, it can only be infinitely close to the maximum.

From the image of the function, it can be seen that when x=0, the value of the function is the smallest, y=0-2t+1=1-2t

About t: No matter what the size of t is, only the height of the image as a whole changes, and it does not affect the shape of the function image, so no matter what value t takes, it is true. So there is no need to discuss the value of t.

Is it t, or am I misunderstood?

1.With respect to y-axis symmetry, the opening is upward, and when x=-2 or 2 is taken, the maximum value is taken, but unfortunately, this is the open interval and can only be.

y<3-2t

2.The function increments at x>0 and decrements by x<0, so the minimum value is y=-2t+1 when x=0

Solution: According to the problem, 5a+2b=3a,a-b=1, a=1 2,b=-1 2. A to the power of 2008 + B to the power of 2009 = 1

x^2-2x-99=0

x^2-2x+1-100=0

x-1)^2=100

x-1 = 10 or x-1 = -10

x = 11 or x = -9

x^2-|x|-2=0

When x is greater than 0.

x^2-x-2=0

x-2)(x+1)=0

x=2,x=-1

Because x is greater than 0, x=2

When x is less than 0.

x^2+x-2=0

x+2)(x-1)=0

x=-2,x=1

Because x is less than 0, x = -2

x²-2x-99=0

x-11)(x+9)=0

x1=11,x2=-9

When x 0, x -x-2 = 0

x-2)(x+1)=0

x1 = 2, x2 = contradict the hypothetical conditions, so discard.

When x 0, x -x-2 = 0

x+2)(x-1)=0

x1 = 1, x2 = contradict the assumptions, so discard.

x=2 or x=-2

x²-2x-99=0

Solution x -2x-99=0

x -2x and 99 are inverses of each other.

x²-2x=99

x²-2x-99 =0

x-11)(x+9)=0

x=11 or -9

x²-|x|-2=0

Solution x -|x|-2=0

x²-|x|and 2 are inverse numbers to each other.

x²-|x|-2=0

x-2)(x+2)=0

x = 2 or -2

The first one writes 99 as +1-100

Category 2 is discussed.

Let A be able to fine x pieces per day, and B can finish Y pieces per day, and the equation can be obtained according to the problem conditions

2x-y=20；

3x+2y=240；

x=40;y=60；

2: If A needs to produce x days, then the number of pieces produced by A is 40x, and B needs to produce (1200-40x) for 60 days.

The required cost is less than 70, and the equation can be obtained:

2*40x + (1200-40x) 60)*x value of the solution equation sphere;