Опростяване на дробни изрази

[Гост]

Здравейте. Ако някой може да ми помогне със следните примери ще съм много благодарен.

Задачите(от подточка А до Е):

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Задача 1
[tex](\frac{2ab}{a^2-b^2}+ \frac {a-b}{2a+2b}). \frac {2a}{a+b} + \frac {a}{b-a}=[/tex]
[tex]=(\frac{2ab}{(a-b)(a+b)}+ \frac {a-b}{2(a+b)}). \frac {2a}{a+b} - \frac {a}{a-b}=[/tex]
[tex]=\frac{4ab+(a-b)^2}{2(a-b)(a+b)}. \frac {2a}{a+b} - \frac {a}{a-b}=[/tex]
[tex]=\frac{4ab+a^2+b^2-2ab}{ \cancel {2 }(a-b)(a+b)}. \frac { \cancel{2}a}{a+b} - \frac {a}{a-b}=[/tex]
[tex]=\frac{\cancel {(a^2+b^2+2ab).}a}{ (a-b) \cancel {(a+b)^2}} - \frac {a}{a-b}=[/tex]
[tex]= \frac {a}{a-b} - \frac {a}{a-b} =0[/tex]

Задача 2
[tex]\frac {x}{x-y} - \frac {x^3-xy^2}{x^2+y^2}. (\frac {x}{(x-y)^2} - \frac {y}{x^2-y^2})=[/tex]
[tex]= \frac {x}{x-y} - \frac {x \cancel {(x^2-y^2)}}{x^2+y^2}. \frac {x(x^2-y^2)-y[(x-y)^2]}{ \cancel {(x^2-y^2)}(x-y)^2}=[/tex]
[tex]= \frac {x}{x-y} - \frac {x }{x^2+y^2}. \frac {x(x-y)(x+y)-y[(x-y)^2]}{(x-y)^2}=[/tex]
[tex]= \frac {x}{x-y} - \frac {x }{x^2+y^2}. \frac {\cancel {(x-y)}[x(x+y)-y(x-y)]}{(x-y)^{\cancel {2}}}=[/tex]
[tex]= \frac {x}{x-y} - \frac {x }{x^2+y^2}. \frac {x^2 \cancel {+xy} \cancel {-yx}+y^2}{x-y}=[/tex]
[tex]= \frac {x}{x-y} - \frac {x }{ \cancel {x^2+y^2}}. \frac { \cancel {x^2 +y^2}}{x-y}=[/tex]
[tex]=\frac {x}{x-y} - \frac {x}{x-y} =0[/tex]

Задача 3
[tex]\frac {3}{3-y} + \frac {y^2+3y}{2y+3}. (\frac {y+3}{y^2-3y} - \frac {y}{y^2-9})=[/tex]
[tex]= \frac {3}{3-y} + \frac {y(y+3)}{2y+3}. (\frac {y+3}{y(y-3)} - \frac {y}{(y-3)(y+3)})=[/tex]
[tex]= \frac {3}{3-y} + \frac { \cancel {y(y+3)}}{2y+3}. \frac {(y+3)^2-y^2}{ \cancel {y(y+3)}(y-3)}=[/tex]
[tex]= \frac {3}{3-y} + \frac { \cancel {y^2 +}6y +9 \cancel {-y^2}}{(2y+3)(y-3)}=[/tex]
[tex]= \frac {3}{3-y} + \frac { 3 \cancel {(2y + 3)} }{\cancel {(2y+3)} (y-3)}=[/tex]
[tex]= \frac {3}{3-y} - \frac {3}{3-y} =0[/tex]

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Задача 4
[tex](\frac {1}{2-x}+ \frac {3x}{x^2-4} - \frac {2}{2+x}):(\frac {4+x^2}{4-x^2}+1)=[/tex]
[tex]= (\frac {1}{2-x}- \frac {3x}{(2-x)(2+x)} - \frac {2}{2+x}): \frac {4 \cancel {+x^2+}4\cancel {-x^2}}{4-x^2} =[/tex]
[tex]= \frac {2+x - 3x -4 +2x}{(2-x)(2+x)}. \frac {4-x^2} {8}=[/tex]
[tex]= \frac {2 \cancel {+x - 3x} -4 \cancel { +2x}}{\cancel {(2-x)(2+x)}}. \frac {\cancel {(2-x)(2+x)}} {8}= - \frac28 = - \frac14[/tex]

Задача 5
[tex]( \frac {2a}{a+2}+ \frac {2a}{6-3a} + \frac {8a}{a^2-4}): \frac {a-4}{a-2}=[/tex]
[tex]=( \frac {2a}{a+2}- \frac {2a}{3(a-2)} + \frac {8a}{(a-2)(a+2)}). \frac {a-2}{a-4}=[/tex]
[tex]= \frac {6a^2-12a-2a^2 - 4a + 24a}{3 \cancel {(a-2) } (a+2)}. \frac { \cancel {a-2}}{a-4}=[/tex]
[tex]= \frac {4a^2+ 8a}{3 (a+2)}. \frac { 1}{a-4}=[/tex]
[tex]= \frac {4a \cancel {(a+2)}}{3 \cancel { (a+2)}}. \frac { 1}{a-4}= \frac {4a}{3a-12}[/tex]

Задача 6
[tex](\frac {1}{a^2-8a+16} + \frac {1}{a^2- 4a}). \frac {2a-8}{a-2} - \frac {1}{a-4} =[/tex]
[tex]= (\frac {1}{(a-4)^2} + \frac {1}{a(a-4)}). \frac {2(a-4)}{a-2} - \frac {1}{a-4} =[/tex]
[tex]= \frac {a+ a - 4}{a(a-4)^{\cancel {2}}}. \frac {2 \cancel {(a-4)}}{a-2} - \frac {1}{a-4} =[/tex]
[tex]= \frac {2a - 4}{a(a-4)}. \frac {2}{a-2} - \frac {1}{a-4} =[/tex]
[tex]= \frac {2 \cancel {(a-2)}}{a(a-4)}. \frac {2}{ \cancel {a-2}} - \frac {1}{a-4} =[/tex]
[tex]= \frac {4}{a(a-4)} - \frac {1}{a-4}= \frac {4-a}{a(a-4)} = - \frac { \cancel {a-4}}{a \cancel {(a-4)}} = - \frac {1}{a}[/tex]

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