13 questions for high school math students come soon

Factorization.

2sin²α-sinαcosα-3cos²α=(2sinα-3cosα)(sinα+cosα)=0

The solution is 2sin -3cos =0, (sin +cos =0 is inconsistent with the known (0, 2), rounded).

And because sin +cos = 1, the solution is sin = 3 13 13, cos = 2 13 13

sin(α+/4)=√2/2sinα+√2/2cosα=√2/2(sinα+cosα)

sin2α+cos2α+1=2sinαcosα+2cos²α=2cosα(sinα+cosα)

The equation sought = 2 (4cos) = 26 4

13.Given (0, 2) and 2sin -sin cos -3cos =0, find the value of [sin( + 4)] (sin2 +cos2 +1).

Solution: From 2sin -sin cos -3cos = (2sin -3cos) (sin +cos )=0, sin +cos = 0, that is, there is tan =-1, (rounded, because it is an acute angle, this formula cannot be established);

and 2sin -3cos = 0, to get tan = 3 2;secα=√(1+tan²α)=√(1+9/4)=√(13/4)=(1/2)√13;

Therefore [sin( + 4)] (sin2 +cos2 +1)=( 2 2)(sin +cos) (2sin cos +2cos).

2/2)(sinα+cosα)/[2cosα(sinα+cosα)]=(√2)/(4cosα)=(√2/4)secα=(√2/4)×(1/2)√13=(1/8)√26.

Turn left|Turn right and answer the question later, pay attention to what is required and define the domain.

8 to the power of 2 to the 3rd is 4, log39 2

So the result is 8

=8-2 3·2, go logarithm on both sides, get.

lg8-2/3=lg2^3-2/3=2lg2, lglog3^9=lglog3-3^2=lg2

The logarithm of the primitive = 2lg2 + lg2 = 3lg2 = lg2 3=lg8 so the primitive = 8

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f(x+1) is an even function on r, f(-x+1) = f(x+1)f(1+x) = f(1-x), the function image is symmetrical with respect to x=1, and if t>0 correctly, then 1+t>1,1-t<1

At x>1, f(1+t)=(1+t-1) 1 (1+t-1)=t + 1 t

At x<1, f(1-t)=f(1+t)=t + 1 t=(1-t-1) 1 (1-t -1).

f(x)=(x-1) 1 (x-1), correct[f(x+1)+2x] x=f(x+1) x +2 function f(x+1) is an even function, f(-x+1)=f(x+1)f(1-x) x=f(1+x) x

f(-x+1) (-x) +2=-f(x+1) x +2f(1+x) x +f(1-x) (-x) =0 function f(1+x) x is an odd function.

m+m=f(1+x) x +2+ f(1-x) (-x) +2=4, correct.

, all correct.

1 2 3

1.Since f(1+x) is an even function of r, there is f(1+x)=f(1-x), so x=1 is the axis of symmetry of the function f(x).

2.Let x1>=0, there is f(1+ x1 )=(1+ x1 -1) 2+ 1 (1+ x1 -1)=x 2+1 x

f(1- x1 )=(1- x1 -1)^2- 1/(1- x1 -1)=x1 ^2+1/ x1 =f(1+x1)

Let x2<0 and have f(1+ x2 ) = (1+ x2 -1) 2- 1 (1+ x2 -1).

x2 ^2-1/x2

f(1- x2 )=(1- x2 -1)^2+ 1/(1- x2 -1)

x2 ^2-1/ x2=f(1 + x2)

So 2 is correct.

3.Let g(x)=f(x+1) x

g(-x)=f(-x+1)/(-x)=f(x+1)/(-x)=-g(x)

So g(x) is an odd function.

Therefore, if g(x) has a maximum and minimum value, then the sum of its maximum and minimum values must be 0

and h(x)=[f(x+1)+2x] x=g(x)+2

So g(x)=h(x)-2

So m-2+n-2=0

So m+n=4

Circle one is about x=-1 symmetry, not 1

f(t)=x is a function with respect to t. x is a constant. Therefore, f(t)=x represents a straight line that crosses the point (0,x) and is parallel to the t-axis.

f(x-1)=x is a function with respect to x-1 and is a parabola.

The two are completely different and have no correlation with each other.

No, the first one is a constant with the parameter x because t is an independent variable and the function does not contain t.

f(x-1)=x 2, substituting x-1 gives f(x)=(x+1) 2

Don't be misleading, the one in f(x) parentheses is called an independent variable, which can be any letter, just a representation, even if there are g(x) these two x doesn't matter, it's just the amount that changes first.

Not the same function.

The dependent variable of the former is t and the function is x 2 and is therefore a constant function.

The latter is a function of x-1.

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