Рационални изрази

[Гост]

Здравейте!Понеже тези задачи ме затрудняват и не знам как да ги реша,може ли да ми помогнете като решите тези задачи?Предварително Ви благодаря!

[ammornil]

(7)[tex]\left(1-\frac{2y}{x}-\frac{y^{2}}{x^{2}}\right)\div (x-y)=\frac{x-y}{x^{2}}[/tex]
Ще се опитаме да преобразуваме лявата страна докато пилучим дясната
[tex]\left(1-\frac{2y}{x}-\frac{y^{2}}{x^{2}}\right)\div (x-y)=\left(\underbrace{1-\frac{2y}{x}-\frac{y^{2}}{x^{2}}}_{\normalsize{x^{2}}}\right)\cdot{\frac{1}{x-y}}=\frac{x^{2}-2xy+y^{2}}{x^{2}}\cdot{\frac{1}{x-y}}=[/tex]

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

[ammornil]

(11) [tex]C=\left[\frac{2}{3x}-\frac{2}{x+1}\cdot \left( \frac{x+1}{3x} -x -1 \right) \right]\div \frac{x-1}{x}=?, \hspace{2em} x=0,5[/tex]

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

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

[tex]C=\underbrace{\left[ \frac{2}{3x} -\frac{2(x+1)(1-3x)}{3x(x+1)} \right]}_{\normalsize{3x(x+1)}}\cdot \frac{x}{x-1}[/tex]

[tex]C=\frac{2(x+1)-2(x+1)(1-3x)}{3x(x+1)}\cdot \frac{x}{x-1}[/tex]

[tex]C=\frac{2(x+1)[1-(1-3x)]}{3x(x+1)}\cdot \frac{x}{x-1}[/tex]

[tex]C=\frac{2(x+1)[1-1+3x)]}{3x(x+1)}\cdot \frac{x}{x-1}[/tex]

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

[tex]C=\frac{2x}{x-1}=\frac{2\cdot{0,5}}{0,5-1}=-\frac{1}{0,5}=-2[/tex]

[ammornil]

(16) [tex]\frac{x^{2}-1}{y}\div \frac{x-1}{y^{2}}=\frac{\cancel{(x-1)}(x+1)}{\cancel{y}}\cdot \frac{y^{\cancel{2}}}{\cancel{x-1}}=y(x+1)=y+xy[/tex]

[ammornil]

(3) [tex]\frac{2a}{a-b}-\frac{x}{b-a}=\frac{2a}{a-b}-\frac{x}{-(b-a)}=\frac{2a}{a-b}-\left(-\frac{x}{a-b}\right)=\frac{2a}{a-b}+\frac{x}{a-b}=\frac{2a+x}{a-b}[/tex]

[ammornil]

(13) [tex]\frac{a}{(x-a)^{2}}+\frac{1}{x-a}-\frac{4}{a-x}=\frac{a}{(x-a)(x+a)}+\frac{1}{x+a}-\left( -\frac{4}{x-a} \right)=\frac{a}{(x-a)(x+a)}+\underbrace{\frac{1}{x-a}+\frac{4}{x-a}}=[/tex]

[tex]=\underbrace{\frac{a}{(x-a)(x+a)}+\frac{5}{x-a}}_{\normalsize{(x-a)(x+a)}}=\frac{1\cdot{a}+(x+a)\cdot{5}}{(x-a)(x+a)}=\frac{a+5x+5a}{(x-a)(x+a)}=\frac{5x+6a}{(x-a)(x+a)}[/tex]