Find the solution to the math problem in the first year of high school, O O thank you

A: Easy to prove that AC is perpendicular to the surface BB1D1D, so it can be proven;

B: You don't need a certificate......

C: The volume of A-BEF is equal to the volume of E-ABB1 minus the volume of F-ABB1, and the difference between the two triangular pyramids is the same bottom, that is, the horizontal component of EF, so the volume of A-BEF can be found, and it is 1 6, so D is selected

Use the sine theorem, the cosine theorem, and do it.

As shown in the figure below:

Why do you always have a bug in the topic today- -Don't be too anxious, kid 1(3a, 4a).

then there is 25a 2=1

a=±1/5

Thus what is sought is (3 5, 4 5).

2.2 Yuan Equation Vertical Notice. into +=1

Solution = -1 = 0

3.Let it be equal to x

Re-order from the meaning of the title (a + into b) * a = 0

then there is 1+x=0

Therefore, the request is -1

4.by the title.

2j-i)*i=2j*i-i 2=2cos60°-1=0 to prove that you can only find these simple questions to do, and the difficult ones don't want to do it.

It's too simple to use trigonometric relationships.

Apply the sine theorem 2r=a sina=b sinb=c sinc

Cosine theorem 2ab*cosc=a2+b2-c2

It's simple.

^sinx+cosx=（1-√3）/2

Both sides of the equation are squared at the same time.

sinx)^2+2sinxcosx+(cosx)^2=(4-2√3)4

Because (sinx) 2+(cosx) 2=1, 2sinxcosx=- 3 2

sin2x=-√3/2

Because x (0, )

So x=60°

tanx=tan60°=√3

Knowing that the real numbers p, q, and r satisfy r2=p+q, r(r-1)=p q, and 0 p 1 q 2, so p and q are the two roots of the equation x2-r2 x+( r2-r)=0, 0 p 1 q 2, so the product of the two roots r2-r 0 can be obtained from the relationship between the roots and the coefficients, and the solution is r 0, or r 1

Let f(x)=x2-r2 x+( r2-r), then from the properties of the quadratic function we get f(1)=1-r2+r2-r 0 , f(2)=4-2r2+r2-r 0

The solution yields r 1, which is solved.

1- 17) 2 r ( 17-1) 2 In summary, 1 r ( 17-1) 2

Hope it helps.

Solving the system of equations 2mx+my=7m and x+y=4 yields (x,y) is the fixed point.

The distance from the straight line to the center of the circle (1,2) is the smallest, the string is the longest, and when the distance is the largest, the chord is the shortest, which is equivalent to solving two inequalities, which can be found by using the distance formula from point to direct.

I just give some ideas, the specific process is calculated slowly, and if you have any questions, please feel free to ask.