Find a math expert to solve sophomore math problems!!

1.Let the equation for the line l be y+3 2=k(x+3), i.e., kx-y+3k-3 2=0

When the string ab is the longest, the line l passes through the center of the circle (0,0), then 3k-3 2=, and the equation for the line l is x-2y=0;

When ab 8, the distance from the center of the circle (0,0) to ab is (5 -4 )=3 from the Pythagorean theorem, and the distance formula from the point to the straight line, 3k-3 2 (k +1)=3, and k=-3 4, then the equation for the straight line l is -3 4x-y+3(-3 4)-3 2=0, i.e., 3x+4y+15=0;

When k does not exist, the line x=3 also satisfies the condition, so the equation for the line l is 3x+4y+15=0, or x=3

2.(What does it mean to be tangent to and tangent to circle c?) The conditions are insufficient, right?).

Circle C: (x-5) y 25

Let the center of the moving circle p(x,y), because the distance between the centers of the two circles is equal to the sum of the radii of the two circles, so the trajectory of the center p of the moving circle is centered on (5,0), and the sum of the radii of the two circles is the circle of radius. (The equation cannot be found).

3.(a x) (x 2) 0, i.e. (x-a) (x 2) 0

Categorical discussion: When a>2, 2 when a<2, and a=2, the solution set is an empty set.

If the chord ab is the longest, then l must pass the origin, so the equation for the straight line l is y=;

If the chord length is 8, then the half-chord length is 4, and the radius is 5, so the distance from the center of the circle to l is 3, so the equation for l is x=-3

a-x) (x-2)>0 is equivalent to (x-a) (x-2)<0;

When a>2, 2 when a<2, and a when a=2, there is no solution.

These are three questions

Question 1 Question 1:

The longest string is the diameter...

Therefore it must pass through the center of the circle.

It's a straight line through the m-point and the origin, you should be able to do it, right?

Second question. The chord length is 8, the radius is 5, and the chord centroid distance is 3

Then the straight line must be tangent to the circle where the equation is x square + y square = 9.

The tangent coordinates are calculated as... I'm too lazy to do the math, you should have taught you this) and then determine the equation for a straight line l based on the tangent and m-point.

Question 2: Did you not express the topic clearly? I don't seem to be able to understand it, it's just a circle C, and the circle tangent to it is everywhere.

Question 3: There are three types of answers.

When a=2, x has no solution.

A>2, 2a<2, a

The longest is the diameter, and after passing through the center of the circle (0,0), the linear equation is calculated by the coordinates of 2 points.

The second one is a straight-line equation to calculate the intersection point, which is solved according to the distance = 8 columns.

The third is to set the coordinates of the center of the circle, and the system of equations of the two equations is listed according to the tangent, which can dissolve the relationship between x and y.

The fourth sub-situation discussion: each is greater than zero or less than zero, and then the situation is divided according to the size of a and 2.

y=[ 0y1=200

I didn't calculate the result, you can do it yourself, it shouldn't be difficult.

Don't transfer us, little sister, you don't have a point.

Look for an equal relationship solution.

There are 11 terms, and the middle term is the 6th term: c(10,5)(-x) 5=-252x 5The sum of the coefficients of all terms is equal to the sum of the coefficients at x=1: (1-1) 10=0 The constant term is equal to 1

So the sum of the coefficients for the rest of the terms is equal to -1

There are 11 items in total, right, so the middle item is definitely item 6. Require the sum of the coefficients of each binomial formula, so that x=1, you can know that all the coefficients add up, =(1-1) 10;

Isn't the sum of the coefficients you finally find in addition to the constant term the sum of the total coefficients minus the constant term 1, so =-1

a(n 1)=2an 3 to the nth power and subtract 3 (n 1) from both sides to the same degree: a(n 1)-3 (n 1)=2an 3 n-3*3 n=2an-2*3 n=2(an-3 n) Therefore, the sequence {an-3 n) is preceding.

Look at the image solution.

The graph is a symmetrical graph with x=a as the axis. and x=a, the minimum value is taken.

So when -1<=a<=1, the minimum value can be taken, f(x)=(x-a) 2-a 2+a-1>=-a 2+a-1

f(x)min (minimum value of the function) = -a 2 + a-1, so that it = 2, the solution is like the problem (note that a has a range.

1<=a<=1）

When a>=1, the function decreases monotonically, and x=1 is the smallest.

When a<-1, increment, x=-1 takes the smallest, and the solution can be brought in, also paying attention to the range of a.

f(x)min represents the minimum value of the function f(x) in a certain interval, and according to the title, this value should be equal to -2;

1: If the positive numbers a and b meet ab a b 3, then the value range of a b is .

Solution: a, b are both integers.

then is obtained by a + b 2ab.

a+b)²≥4ab

i.e. ab (a+b) 4

AB a b 3

then a b 3 (a+b) 4

Let a+b=t>0

then t 4 - t -3 0

The solution is t 6 or t -2<0 (rounded), i.e. a+b 6

2: Mix a kilogram of sugar with water to prepare a kilogram of sugar water (b is greater than a and greater than zero), then its concentration is? If you add m kg of sugar, (m is greater than zero), the sugar water is sweeter, and a common inequality is extracted according to the common sense of life.

Solution: (1) The concentration l is l=a b x 100% (2) (a+m) (b+m)> a b (m>0), that is, the numerator and denominator of a fraction are always greater than the value of that fraction if you are interested!

3: The point p(3,0) is used as a straight line l, so that the midpoint of the line segment intercepted by two intersecting lines 2x y 2 0 and x y 3 0 is exactly bisected by p, then the equation for the straight line l is ??? Thanks, urgent.

Solution: Method 1.

Let y=k(x-3).

Let the found line intersect 2x-y-2=0 at (a,2a-2) and x+y+3=0 at (b,-b-3), so the union has 2a-b-5=0

a+b=6 solves a=11 3

b = 7 3 so substituting k = 8 to get the straight line found is y = 8x-24 method two straight line system equation through p point is: y = k (x-3) and equations 2 x - y - 2 = 0 and x + y + 3 = 0 respectively, the obtained intersection abscissa is:

3k-2) (k-2) and (3k-3) (k+1) since p is the midpoint of these two intersections.

3k-2) (k-2) + (3k-3) (k+1) = 3 2 passes are sorted out, and k = 8 is obtained

So the equation is: y=8(x-3) =8x-24 is the same method in the end.

Question 1: a+b=ab-3

Because a 2+b 2+2ab 2ab+2ab=4ab, i.e. ab (a+b) 2 4

So a+b (a+b) 2 4-3

Let a+b=t

t≤t^2/4-3

t^2-4t-12≥0

t≤-2 or t≥6

Obviously a+b>0

So x=a+b 6

(m-3)(n-1)=mn-3n-m+3=m(n-1)-3n+3=3m(n-1)-3n=0, m=3/(n-1)=3+3/(n-1)m+n=3+3/(n-1)+n=4+3/(n-1)+(n-1)>=4+2√3

The condition for the equal sign to be true is 3 (n-1)=n-1, that is, the minimum value of n=1+ 3m+n is 4+2 3

Using the equation, it is found that m is equal to n minus 3/1 plus 3, so m+n=3+(3 (n-1))+n=4+((3 (n-1))+n-1 The last two terms are greater than or equal to 2 times the root number 3, so m+n is greater than or equal to 4+2 times the root number 3

Have you learned basic inequalities?