Can anyone help me solve a few math problems?

1, when x=1, bring in, and solve f(3)=-1

2. Because g(x)=2 x, and g(a)g(b)=16, a+b=4, log2(a+b)=log2(4)=2

3. Classification discussion, if x>1, the equation is constant, if 0y1, when a=1 y2=y1, when a>1, y1>y2

5，lg(10^3-10^2)=lg900=lg9*100=lg9+lg100=lg9+2

6, log8 9=log2 3 3 2=(2 3) log2 3 up and down about log2 3 to get 2 3

I hope you are satisfied, thank you.

1.Let 2x+1=3 x=1 bring in f(3)=-12Since g(x)=2 x, and g(a)g(b)=16, a+b=4, log2(a+b)=log2(4)=2

3.If x>1, the equation is constant, if 0y1, y2=y1 when a=1, and y1 when a>1, y1>y2

6.Bottom change formula. log8 9=lg9/lg8=lg3^2/lg2^32lg3/3lg2=2/3 log2 3

The reason is clear enough.

1--- when x=1, bring in -1

2--- when x=2 and a=b=2, the answer is 2

Have you ever copied the wrong question, the first letter of the two sets is a, and two equations can't find three unknowns, so there will be an infinite number of solutions.

Analysis: This is the basic definition of a collection of examiners.

Solution: According to the heterogeneity of sets, the elements in a and b must not be equal, i.e.:

a≠a+b≠a+2b

a≠ac≠ac²

Therefore: b≠0, a≠0

When the set is equal, the elements in the set must be equal, and according to the title can only be:

a+b=ac

a+2b=ac²

or a+b=ac

a+2b=ac

Whereas: a+b=ac

a+2b=ac²

Solution: b=ac-a

a+2b=a+2ac-2a=ac, i.e., 2c-1=c, obtain: c=1 verification: at this time: a= contradicts the set definition, c=1 rounds a+b=ac

a+2b=ac

Solution: b=(ac-a) 2, get:

a+b=a+(ac-a) 2=ac, i.e., 2c -c-1=0, the solution is:

c=-1 2 or c=1 (rounded).

Validation: When c=-1 2, the set condition is satisfied.

Therefore: c=-1 2

1, because the three sides are integers, so from the area formula s=absin(c) 2, we know that there must be one of them is 60 degrees or 120 degrees, and 120 cannot make a, b, and c into equal differences, so we know that there must be a corner of 60 degrees. You might as well let this angle be c, and the area formula before the substitution can be obtained: a*b=40

1) and a+b+c=20(2), from (1) can obtain the combination of integers:

1*40=40;Off topic.

2*20=40;Off topic.

4*10=40;c=6, these three sides do not form a triangle, which is not in line with the topic, 5*8=40; c=7, that's it.

2. You may wish to let x=sina, and the original equation is: (x-m) 2+m(2-m)>0, from the title: m(2-m)>0, 03 is obtained, and 2rsana=a

1,f(condition,f(,f(,f(,f(,f(,,f(,

2,f(1-2a)<-f(1-a),f(1-2a)a-1 function decreases, and the solution is a<2 shed 3,-1<1-2a<1,-1 The rest of the gods are also clever.

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Solution: Let the coordinates of point p be (x,0).

Then the slope of the straight line AP k1=(0-3) (x+2)=-3 (x+2)=tan

The slope of the straight line Pb k2=(7-0) (5-x)=7 (5-x)=tan( -=-tan

Because the light rays are reflected, the angle of incidence = the angle of reflection.

So k1=-k2

i.e. -3 (x+2)=-7 (5-x).

Solution: x=1 10.

So the coordinates of point p are (1 10, 0).

The reflection is axisymmetric.

Find a point of symmetry about the x-axis, such as the symmetry point a'(-2, -3) of a(-2,-3).

Find the intersection point of the line a'b and the x-axis is the p point.

The symmetry point of point A(-2,3) with respect to the x-axis is point C(-2,-3), the slope of the line BC is k=10 7, and the equation for the line BC is:

y-7=k(x-5), when y=0, x=1 10, so the coordinates of the point p are (1 10,0).

Good luck with your studies!

The symmetry point a' of point A is made through the x-axis, and the link is a'b, then a'The intersection of b and the x-axis is the point p, known a(-2,3), so a'(-2, -3), there is b(5,7) and the line a'b is y=kx+b put a'(-2,-3) b(5,7) is substituted to find y=10 7x-1 7

When y=0, it is the p-point coordinate (1, 10, 0).

The symmetry point of the incident point A(-2,3) on the x-axis is A1(-2,-3), then the point A1 should be on the straight line where the reflected ray is located, that is, a1,p,b three points are collinear, that is, ka1p=kpb,3 (x+2 )=7 (5-x), and the solution is x= , i.e., p( ,0).

The equation for the reflected rays passes through the points (-2, -3), b(5, 7) is 10x-7y-1 = 0 coordinates of the point p (1 10, 0).