Two math problems in the second year of high school, seeking masters!! Velocity!!!

1. The slope of the tangent x+y-1=0 is -1

The slope of the normal equation of the circle made by the tangent point (2,-1) is 1y -(1) = x - 2 , that is, the center of the circle of y = x - 3 is the intersection of the straight line 2x+y=0 and y = x - 3, and (1,-2) is obtained

The radius is the distance between the points (2,-1) and (1,-2).

The square of the radius r2 = 2

The equation for the circle is: (x - 1) 2 + y + 2) 2 = 22, let m(x1,y1), n(x2,y2), and the equation for the line l be x=10 3 or y=k(x-10 3), and the distance from m,n to the straight line x=2 is d1 and d2 respectively

1) If the equation for the straight line l is x=10 3, there is x1=x2=10 3, d1=d2=2-10 3=-4 3

d1+d2=2(4 3)≠10 3, 2) If the equation of the straight line l is y=k(x-10 3), there is x 2+4k 2(x-10 3) 2-4=0 to calculate the value of k.

Solution: k=?

The equation for the straight line l is: y=kx+b.

(1)y=[50-3(x-6)]x-115=68x-3x*x-115 defines the domain 50-3(x-6)>0 Because x is an integer, x<22 defines the domain as [6,22].

2) Find the maximum value of a quadratic equation.

y=【50-3（x-6）】x-115=68x-3x*x-115a=-3 b=68 c=-115

b 2a= so when x=11 takes the maximum value y=270

1. If you don't start at the base, according to the principle of the shortest straight line between two points, A and B go directly to C with the shortest distance, so A goes directly to the east, and B, with B as the origin as the coordinate axis, you can know the tan angle YBC = 2 3 (please calculate how many degrees yourself, assuming x degrees. So b is x degrees east-northerly.

2. If you want to start at the base, then AB should first walk back to the base and walk a total of 4 meters, and then start at point O, take point O as the origin as the coordinate axis, the principle is the same, and get the tan angle AOC = 5 2 (please calculate the degree yourself, assuming x. So go at point o north-east x degrees. The shortest distance is 4+ root number 29

The title is weird.

Here are some examples! Check it out! Won't ask me again!

1.Let the first term a, the common ratio q

then a 3 * q 12 = 8

So a 2 * q 8 = 4

a2a8=a^2*q^8=4

Therefore a9 is a constant.

So a1a17=(a9) 2....a8a10=(a9) 2t17 is constant.

6.Directly exploit the formula:

sin(α+sinαcosβ+cosαsinβ∴sin(x+π/3)=sinxcos(π/3)+cosxsin(π/3)

1/2)sinx+(√3/2)cosxf(x)=2cosxsin(x+π/3)-√3/2=2cosx[(1/2)sinx+(√3/2)cosx ]-3/2=sin2x+√3(cosx)^2-√3/2

锛宎2a8=(a1q 4) 2=2 =4 甯告暟 t17=a1 17q 136=(a1q 8 ) 17=甯告暟 |3/2cosx)-|3/2=sinxcosx |3cosx^2-|3/2=1/2sin2x |3 2cos2x=sin(2x scandium 3).

（1/a^2 -1)(1/b^2 -1)=（1-a^2 )(1-b^2 )/a^2b^2=[1-(1-b)^2 ](1-b^2 )/1-b)^2b^2=(b^2-b-2)/(b^2-b)=1-2/(b^2-b)=1-2/[(b-1/2)^2-1/4]

When [(b-1 2) 2-1 4] is the smallest, it is the minimum value of (1 a 2 -1) (1 b 2 -1). So when b=1 2, there is a minimum value = 9

It is obtained from 2sinacosc=sinb=sin(a+c)=sinacosc+cosasinc.

sinacosc=cosasinc

tana = tanc, so angle a = angle c, a = c

a+b）²-c²=3ac=3c²

a+b）²=4c²=4a²

So a+b=2a, b=a

Therefore, the triangle ABC is an equilateral triangle.