Question

Представьте в виде дроби:
А) 4/у+2- 3/у-2+ 12/у²-4
Б) a/a-6-3/a+6+a²/36-a²
B) x²/(x-y)²-x+y/2x-2y
Г) b/(a-b)²-a+b/b²-ab.
Помогите решить!

Answer 1

$1)\dfrac{4}{y+2}-\dfrac{3}{y-2} +\dfrac{12}{y^{2}-4 } =\dfrac{4}{y+2}-\dfrac{3}{y-2} +\dfrac{12}{(y+2)(y-2) }=\\\\=\dfrac{4\cdot(y-2)-3\cdot(y+2)+12}{(y+2)(y-2)}=\dfrac{4y-8-3y-6+12}{(y-2)(y+2)}=\\\\=\dfrac{y-2}{(y+2)(y-2)}=\boxed{\dfrac{1}{y+2}}$

$2)\dfrac{a}{a-6}-\dfrac{3}{a+6}+\dfrac{a^{2} }{36-a^{2} } =\dfrac{a}{a-6}-\dfrac{3}{a+6}-\dfrac{a^{2} }{a^{2}-36 } =\\\\=\dfrac{a}{a-6}-\dfrac{3}{a+6}-\dfrac{a^{2} }{(a-6)(a+6) } =\dfrac{a\cdot(a+6)-3\cdot(a-6)-a^{2} }{(a-6)(a+6)}=\\\\=\dfrac{a^{2}+6a-3a+18-a^{2}}{(a-6)(a+6)}=\dfrac{3a+18}{(a-6)(a+6)} =\dfrac{3(a+6)}{(a-6)(a+6)}=\boxed{\dfrac{3}{a-6} }$

$3)\dfrac{x^{2} }{(x-y)^{2} } -\dfrac{x+y}{2x-2y}=\dfrac{x^{2} }{(x-y)^{2} }-\dfrac{x+y}{2(x-y)}=\dfrac{2x^{2}-(x+y)(x-y) }{2(x-y)^{2} } =\\\\=\dfrac{2x^{2}-x^{2}+y^{2}}{2(x-y)^{2} }=\boxed{\dfrac{x^{2}+y^{2}}{2(x-y)^{2} } }$

$4)\dfrac{b}{(a-b)^{2} }-\dfrac{a+b}{b^{2}-ab }=\dfrac{b}{(a-b)^{2} }+\dfrac{a+b}{b(a-b) }=\\\\=\dfrac{b^{2}+(a+b)(a-b) }{b(a-b)^{2} }=\dfrac{ b^{2}+a^{2}-b^{2}}{b(a-b)^{2} }=\boxed{\dfrac{a^{2} }{b(a-b)^{2} }}$