используя тригонометрические формулы и преобразования
получаем:
что и требовалось доказать
((ctgx)^x)`= (ctgx)^x*ln(ctgx)*(ctgx)`= (ctgx)^x)*ln(ctgx)*(-1/sin^2x)=
= -((ctgx)^x*ln(ctgx))/sin^2x
(x-2)²(x-3)+(x-2)(x-3)²= (x-2)(x-3)(x-2+x-3)= (x-2)(x-3)(2x-5)
3m(m+2n)-2n(m+2n)²= (m+2n)(3m-2n(m+2n)) = (m+2n)(3m-2mn-4n²))
(а+3)/(а-1) + (а+5)/(2-2а) =
= (а+3)/(а-1) + (а+5)/ -2(а-1) =
= (а+3)/(а-1) - (а+5)/2(а-1)=
= (2(а+3) - (а+5))/ 2(а-1)=
= (2а+6 -а -5) /2(а-1)=
= (а+1)/2(а-1) =
= (а+1)/(2а-2)
4m/(m²-n²) - 4/(m+n) =
= (4m - 4(m-n) ) / (m+n)(m-n) =
= (4m - 4m +4n) / (m+n)(m-n) =
= 4n/(m²-n²)
(2x-y)/(2x+y) + y/(4x+2y) =
= (2x-y)/(2x+y) + y/2(2x+y)=
= ( 2(2x-y) +y) / 2(2x+y)=
= (4x-2y+y) / 2(2x+y)=
= (4x-y)/ 2(2x+y) =
= (4x-y) / (4x+2y)
(x²+y²)/(x²-y²) - (x-y)/(x+y) =
=(x²+y² - (x-y)(x-y) ) / (x-y)(x+y) =
= (x²+y² - (x² -2xy +y²)) / (x-y)(x+y) =
= (x² +y²-x²+2xy -y²) / (x-y)(x+y) =
= 2xy / (x²-y²)
4/(a²-9) -2/(a²+3a) =
= 4/(a-3)(a+3) - 2/ a(a+3) =
= (4a -2(a-3)) / (a(a-3)(a+3) )=
= (4a -2a+6) / a(a-3)(a+3) =
= (2a+6) / a(a-3)(a+3) =
= 2(a+3) / a(a-3)(a+3) =
= 2/ a(a-3)
1/(2m -4) + m/(4-m²) =
= 1/ 2(m-2) - m/(m-2)(m+2) =
= (m+2 - 2m) / 2(m-2)(m+2) =
= (-m+2) / 2(m-2)(m+2) =
= - (m-2)/ 2(m-2)(m+2) =
= - 1/ 2(m+2) =
= - 1/(2m+4)
(4x²+4)/(16x²-9) - (x²+1)/(4x² -3x) =
= (4x²+4) / (4x-3)(4x+3) - (x²+1) / x(4x-3) =
= (x(4x²+4) - (x²+1)(4x -3)) / ( x(4x-3)(4x+3) )=
= (4x³ +4x - (4x³ -3x² +4x-3)) / (x(4x-3)(4x+3)) =
= ( -3x² +3 ) / (x(4x-3)(4x+3)) =
= - (3x² -3)/ (x (16x² -9 )) =
= - (3x² - 3) / (16x³ -9x)
b/(3b-1) + (3b² -1)/ (1-9b²) =
= b/(3b-1) - (3b² -1) / (3b-1)(3b+1) =
= (b(3b+1) - (3b² - 1) ) / (3b-1)(3b+1) =
= (3b² + b - 3b² + 1) / (3b-1)(3b+1) =
= (b+1) /(9b² - 1)