t(r+ 1)=c6(r)(x)^(6-r)(- 1/√x)^r

=(- 1)^rC6(r)x^(6-r-r/2)

Is a constant term, so

6-r-r/2=0, then r=4.

So (-1) RC6 (r)

=(- 1)^4*C6(4)= 15

So the constant term in the sixth power expansion of binomial (x-1√ x) is 15.

2.sinx+cosx =- 1, where both sides are squared at the same time:

1+2sinxcosx= 1

sin2x=0

{x|x=kπ/2 k is an integer}

3.2 C2+3c 2+4c 2+……+nC2 = 3c 3+3c 2+4c 2+……+nC2

= 4c3+4c2+...+nc2 (keep pushing)

=nC3=n(n- 1)(n-2)/6

The numerator and denominator are divided by n 3 at the same time, and the limit is 1/6.

4. If point P(-2, √3) is on the terminal side of angle α, then sinα=y/r=√3/√7, cosα=x/r=-2/√7,

sin(α+π/3)

= Sina cosπ/3+cosa sinπ/3 =(√3/√7)*( 1/2)+(-2/√7)*(√3/2)=(√3-2√3)/(2√7)=-√3/(2√7)