Solving math is a master of solving, and a master of mathematics is advancing

<> compare to see that the coefficient is.

Requires x(x-2 7) 7 in the equation where x 4 is a factor.

That is, find the coefficient of x 3 in the equation (x-2 7) 7.

is the coefficient of the fifth term in the formula, i.e., (-2 7) 48xc7(4) The 7 after (4) is the subscript and the 4 in parentheses is the superscript.

So the result is 80 343

tr+1=x[c(r 7)x^(7-r)*(2/7)^r]=c(r 7)x^(8-r)*(2/7)^r

Known : 8-r=4 r=4

So the term x 4 is c(4 7) x (8-4)*(2 7) 4=35*x 4*16 2401=80 343x 4

The coefficient is 80 343

Will the ones upstairs count?! Don't mistake people, whether it's good or not.

x 4 and removing an x is x 3 for the fifth term of this binomial c(4 7) and the first term is x 7*x

I don't even understand the binomial theorem, so what nonsense do you come out of?

The binomial theorem, the coefficient of the third term of x, which is the fifth term, is the required coefficient.

Because there is x in front of it, it needs x 3 in the back, 7c3 (or 3c7?). )*2 7) 4=80 343 I can't calculate it, you can check it.

= (2/2 power of 3) (2/3 power of 3) (3/3 power 3) = 2/23 power of 3.

Root number 3 = 3 (1 2) power.

The third root number 9 = 9 (1 3) power = 3 power (2 3) power 4 root number 27 = 27 power (1 4) power = 3 (3 4) power addition = 3 (1 2 + 2 3 + 3 4) power of 3.

3 (23 to 12) power.

Hope you understand.

1 solution: Let the hyperbola be: y 2 a 2

x^2/b^2=1

Questioned: e=c a=2 3;Let's punch letter A 2

b^2=c^2

and 18 a 2

4/b^2=1

Solution: a= 6, b= 2

The hyperbolic equation is: y 2 6

x^2/2=1

Easy to verify: The tanwheel focus does not hold on the x-axis. ）

2 Solution: p and q and non-q are false propositions at the same time.

p false q true. x^2-x<6

i.e.: -2

l1/sin（b-x）=l2/sin（（a+x）-（b-x））

l1sin（2x+a-b）+l2sin（x-b）=0

Then the known substitution can be solved.

This is an indefinite system of equations. Exponents b1,b2 are decimals, a1,a2 should be limited to positive integers.

When a1=1, c1=1, a2 b2=, a2 is an integer greater than 1, and b2=

Let 2<=a1<=a2, then.

a1 b1*a2 b2>=2 (b1+b2)=2 ,2 40>[(c1+c2) 2] 2=>=cc1=2 ,contradiction.

In summary, a1=1, c1=1, c2=, a2 is an integer greater than 1, b2=b1=

Because the difference between the distance from p to f(2,0) and the distance from p to the y axis is equal to 2, the distance from p to f is equal to the distance from p to the straight line x=-2, so the trajectory of the center p is a parabola focused on f(2,0), so the equation for trajectory c is y = 8x

The second one will not seek to adopt o( o