Опростете изразите.

[namelessgem]

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


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

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

Благодаря предварително. : )

[Добромир Глухаров]

[tex]1+(x-1):\left(x+\frac{1}{x-2}\right)=1+(x-1):\frac{x(x-2)+1}{x-2}=1+(x-1):\frac{x^2-2x+1}{x-2}=\\=1+(x-1)\cdot\frac{x-2}{(x-1)^2}=1+\frac{x-2}{x-1}=\frac{x-1+x-2}{x-1}=\frac{2x-3}{x-1}[/tex]

[Добромир Глухаров]

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

[tex]=\frac{x}{1-x}\cdot\frac{1-x}{\cancel{x}-x^2\cancel{-1}+2x^2\cancel{+1}\cancel{-x}}=\frac{x}{\cancel{1-x}}\cdot\frac{\cancel{1-x}}{x^2}=\frac{1}{x}[/tex]

[Добромир Глухаров]

[tex]\left(\frac{1}{a-1}\cdot\frac{1}{a+1}-1\right)\cdot\frac{a^2-1}{a}=\left(\frac{1}{a^2-1}-1\right)\cdot\frac{a^2-1}{a}=\\=\frac{1-(a^2-1)}{\cancel{a^2-1}}\cdot\frac{\cancel{a^2-1}}{a}=\frac{2-a^2}{a}[/tex]

[namelessgem]

[tex]\left(\frac{1}{a-1}\cdot\frac{1}{a+1}-1\right)\cdot\frac{a^2-1}{a}=\left(\frac{1}{a^2-1}-1\right)\cdot\frac{a^2-1}{a}=\\=\frac{1-(a^2-1)}{\cancel{a^2-1}}\cdot\frac{\cancel{a^2-1}}{a}=\frac{2-a^2}{a}[/tex]

Във първите скоби по условие е изваждане, а не умножение.

[Добромир Глухаров]

[tex]\left(\frac{1}{a-1}-\frac{1}{a+1}-1\right)\cdot\frac{a^2-1}{a}=\frac{a+1-(a-1)-(a^2-1)}{a^2-1}\cdot\frac{a^2-1}{a}=\\=\frac{\cancel{a}+1\cancel{-a}+1-a^2+1}{a^2-1}\cdot\frac{a^2-1}{a}=\frac{3-a^2}{\cancel{a^2-1}}\cdot\frac{\cancel{a^2-1}}{a}=\frac{3-a^2}{a}[/tex]