[[,=2((15)/(2)-(2)/(2)],[,=(26+7)/(2)=]],[" 6."quad m-(m-1)/(2)=1-(m-2)/(3)]

If m+(1)/(m-2)=4 then (m-2)^(2)+(1)/((m-2)^(2))=

m(m-1)(m-2)...3.2.1=

The area of the triangle formed by the lines y=m_(1x)+c_(1),y=m_(2),x+c_(2) and x=0 is (1)/(2)((c_(1)+c_(2))^(2))/(|m_(1)-m_(2)|) b.(1)/(2)((c_(1)-c_(2))^(2))/(|m_(1)+m_(2)|) c.(1)/(2)((c_(1)-c_(2))^(2))/(|m_(1)-m_(2)|) d.((c_(1)-c_(2))^(2))/(|m_(1)-m_(2)|)

Fiind m, if the lines (1-x)/(3)= ( 7y-14)/(2m ) = (z-3) /(2) and ( 7-7x)/(3m) = (y-5)/(1) = (6-z)/(5) are at right angles.

2m-(1)/(2m)=2 , m!=0, in m^(2)+(1)/(16m^(2)) =?

sum_(r=0)^n(-1)^r^n C_r[1/(2^r)+3/(2^(2r))+7/(2^(3r))+(15)/(2^(4r))+ u ptomt e r m s]= (2^(m n)-1)/(2^(m n)(2^n-1))

sum_(r=0)^n(-1)^r .^n C_r[1/(2^r)+3/(2^(2r))+7/(2^(3r))+(15)/(2^(4r))+ .....mt e r m s]= (2^(m n)-1)/(2^(m n)(2^n-1))