Senior high school math exercises

Suppose the center of the circle is (x0, 3x0).

Then the radius is 3x 0(x0 & gt；; 0)

The distance from the center of the circle to y=x is |x0-3x0|/√2=√2x0.

Then (2 √ 7/2) 2 = (3x0) 2-(√ 2x0) 2.

X0= 1。

So the center of the circle is (1, 3) and the radius is 3.

So (y-3) 2+(x- 1) 2 = 9.

Po 2-om 2 = op 2 (let P(x 1, y 1).

Then 2x 1+6y 1= 1.

Then find the minimum value of √ (x12+y12).

Various methods can be used.

For example, eliminating one yuan. ..

And Cauchy ..

Number-shape combination

I write the answer directly.

The answer is √ 10/20.

2.f(x)=-f(-x)

Then LG (1+ax)/(1+2x) =-LG (1-ax)/(1-2x).

If a 2 = 4, then a=-2 or a=2 (discard)

So f (x) = LG (1-2x)/(1+2x).

And (1-2x)/( 1+2x) >: 0.

Yes-1/2

Then-b > -1/2 and b

So 0

f(x)= LG( 1-2x)/( 1+2x)= LG {- 1+[2/( 1+2x)]}

As 1+2x >: 0, 1+2x does not increase with the increase of x.

2/( 1+2x) will not reduce ... so-b.

& ltb, at this time f(x) is decreasing,

When 1+2x

f(x+ 1)-f(x)=[(x+ 1)(x+2)[35-2(x+ 1)]-x(x+ 1)(35-2x)]/ 150

=(x+ 1)(33-x)/25

f( 1)=g( 1)=7/30

g(x)= g( 1)+( 1+ 1)(33- 1)/25+(2+ 1)(33-2)/25......(x+ 1)(33-x)/25

....

Can you write the title clearly?

Think there's something wrong.