Опростяване & доказване на тъждества.

[Lilly.]

1. Опростете изразите:

[tex](x^2 - 1)(\frac{1}{x - 1} - \frac{1}{x + 1} - 1);[/tex]

[tex](\frac{a^2}{b^2} + \frac{b}{a}) : (\frac{1}{a} - \frac{1}{b} + \frac{a}{b^2});[/tex]

[tex](\frac{x}{x - 2} - \frac{2}{x + 2}) : (\frac{x+2}{2} - \frac{x - 2}{x});[/tex]

[tex](\frac{a}{x - a} - \frac{a}{x + a}) . \frac{x^2 + 2ax + a^2}{2a^2}.[/tex]

2. Докажете тъждествата:

[tex](x - \frac{1}{x}) : (\frac{1}{x} + 1);[/tex]

[tex]\frac{ab}{a + b} ( \frac{1}{a} - \frac{1}{b}) = \frac{b - a}{b + a};[/tex]

[tex](1 - \frac{2y}{x} + \frac{y^2}{x^2}) : (x-y) = \frac{x - y}{x^2};[/tex]

[tex](x + 1 - \frac{1}{ 1 - x}) : (x - \frac{x^2}{x - 1}) = - x.[/tex]

Благодаря предварително! ( :

[ammornil]

1. Опростете изразите:

[tex](x^2 - 1)(\frac{1}{x - 1} - \frac{1}{x + 1} - 1)= \\
=\cancel{(x^{2}-1)}.\frac{x+1-(x-1)-(x^{2}-1)}{\cancel{x^{2}-1}}=\cancel{x}+1\cancel{-x}+1-x^{2}+1=3-x^{2}[/tex]

[tex](\frac{a^2}{b^2} + \frac{b}{a}) : (\frac{1}{a} - \frac{1}{b} + \frac{a}{b^2})= \\
=\frac{a^{3}+b^{3}}{a.b^{2}}\hspace{1}:\hspace{1} \frac{b^{2}-a.b+a^{2}}{a.b^{2}}=\frac{(a+b).\cancel{(a^{2}-a.b+b^{2})}}{\cancel{a.b^{2}}}\hspace{1}.\hspace{1} \frac{\cancel{a.b^{2}}}{\cancel{a^{2}-a.b+b^{2}}}=a+b[/tex]

[tex](\frac{x}{x - 2} - \frac{2}{x + 2}) : (\frac{x+2}{2} - \frac{x - 2}{x})= \\
=\frac{x.(x+2)-2.(x-2)}{(x+2).(x-2)} \hspace{1} : \hspace{1} \frac{x.(x+2)-2.(x-2)}{2.x}=\frac{\cancel{x.(x+2)-2.(x-2)}}{(x+2).(x-2)} \hspace{1} . \hspace{1} \frac{2.x}{\cancel{x.(x+2)-2.(x-2)}}=\frac{2.x}{(x+2).(x-2)}[/tex]

[tex](\frac{a}{x - a} - \frac{a}{x + a}) . \frac{x^2 + 2ax + a^2}{2a^2}= \\
=\frac{a.(x+a)-a.(x-a)}{(x-a).(x+a)}\hspace{1}.\hspace{1}\frac{(x+a)^{2}}{2.a^{2}}=\frac{a.(x+a-x+a)}{(x-a).(x+a)}\hspace{1}.\hspace{1}\frac{(x+a)^{2}}{2.a^{2}}= \\
\frac{\cancel{a^{2}}}{(x-a).\cancel{(x+a)}}\hspace{1}.\hspace{1}\frac{(x+a)^{\cancel{2}}}{2.\cancel{a^{2}}}=\frac{x+a}{2.(x-a)}[/tex]

[ammornil]

2. Докажете тъждествата:

[tex](x - \frac{1}{x}) : (\frac{1}{x} + 1) \hspace{6} \hspace{2}[/tex] това е израз а не равенство...
..::
[tex]\frac{ab}{a + b} ( \frac{1}{a} - \frac{1}{b}) = \frac{b - a}{b + a}; \\
\frac{a.b}{a + b}. \left( \frac{1}{a} - \frac{1}{b} \right) =\frac{a.b}{a + b}. \frac{b-a}{a.b}=\frac{b-a}{b+a} \hspace{24} (QeD).[/tex]
..::
[tex](1 - \frac{2y}{x} + \frac{y^2}{x^2}) : (x-y) = \frac{x - y}{x^2}; \\
(1 - \frac{2.y}{x} + \frac{y^2}{x^2}) : (x-y) =\frac{x^{2}-2.x.y+y^{2}}{x^{2}}.\frac{1}{x-y}=\frac{(x-y)^{\cancel{2}}}{x^{2}.\cancel{(x-y)}}=\frac{x-y}{x^{2}} \hspace{24} (QeD).[/tex]
..::
[tex](x + 1 - \frac{1}{ 1 - x}) : (x - \frac{x^2}{x - 1}) = - x \\
(x + 1 - \frac{1}{ 1 - x}) : (x - \frac{x^2}{x - 1})=\frac{(1+x).(1-x)-1}{1-x}\hspace{2}:\hspace{2}\frac{x.(x-1)-x^{2}}{x-1}=\frac{\cancel{1}-x^{2}\cancel{-1}}{-(x-1)}\hspace{2}.\hspace{2}\frac{x-1}{\cancel{x^{2}}-x\cancel{-x^{2}}}= \\
=\frac{\cancel{-1}.x^{2}}{\cancel{-1}.(x-1)}.\frac{x-1}{-x}=\frac{x^{\cancel{2}}}{\cancel{x-1}} \hspace{2}.\hspace{2} \frac{\cancel{x-1}}{-1.\cancel{x}}=-x \hspace{24} (QeD).[/tex]