Действия с рационални дроби (трудни знадачи)

[Гост]

Здравейте! Ето тези задачи са ми адски трудни и нито една от тях не можах да реша:

1. [tex]\frac{[-3(x-1)]^{3}}{[2ax]^{3}} : \frac{3x-3}{2a^{2}x}[/tex]

2. [tex]\frac{3p^{2}mq}{5d^{2}b^{3}} . \frac{9abc}{8x^{2}y^{2}} : \frac{3a^{2}b^{2}c^{3}}{15pxy}[/tex]

3. [tex]\frac{(x+y)^{3}}{xy-y^{2}} : \frac{[-xy+y^{2}]}{[(x-y)^{2}]}[/tex]

4. [tex]\frac{3a^{2}+3ab+3b^{2}}{5a+5b} . \frac{10a^{2}-10b^{2}}{9a^{3}-9b{3}}[/tex]

[ammornil]

[tex]\frac{[-3.(x-1)]^{3}}{[2.a.x]^{3}} : \frac{3.x-3}{2.a^{2}.x}=\frac{-3^{3}.(x-1)^{3}}{2^{3}.a^{3}.x^{3}}.\frac{2a^{2}x}{3.(x-1)}=[/tex]
[tex]-\frac{\cancel{2}.3^{2 \\ \cancel{3}}.\cancel{a^{2}}.\cancel{x}.(x-1)^{2 \\ \cancel{3}}}{2^{2 \\ \cancel{3} }.\cancel{3}.a^{\cancel{3}}.x^{2 \\ \cancel{3}}.\cancel{(x-1)}}=[/tex]
[tex]-\frac{9.(x-1)^{2}}{4.a.x^{2}}=-\frac{1}{a}.\left[\frac{3.(x-1)}{2.x} \right]^{2}[/tex]

[ammornil]

[tex]\frac{3.p^{2}.m.q}{5.d^{2}.b^{3}} . \frac{9.a.b.c}{8.x^{2}.y^{2}} : \frac{\cancel{3}.a^{2}.b^{2}.c^{3}}{\cancel{15}.p.x.y \\ 5}=[/tex]
[tex]\frac{3^{3}.\cancel{a}.\cancel{b}.\cancel{c}.m.p^{2}.q}{2^{3}.\cancel{5}.b^{ 2 \\ \cancel{3} }.d^{2}.x^{\cancel{2}}.y^{\cancel{2}}}\hspace{2}.\hspace{2}\frac{\cancel{5}.p.\cancel{x}.\cancel{y}}{a^{\cancel{2}}.b^{2}.c^{ 2 \\ \cancel{3}}}=\frac{27.m.p^{3}.q}{8.a.b^{4}.c^{2}.d^{2}.x.y}[/tex]

[ammornil]

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

[ammornil]

[tex]\frac{3.a^{2}+3.a.b+3.b^{2}}{5.a+5.b} \hspace{2} . \hspace{2} \frac{10.a^{2}-10.b^{2}}{9.a^{3}-9.b{3}}=\frac{3.(a^{2}+a.b+b^{2})}{5.(a+b)} \hspace{2} . \hspace{2} \frac{10.(a^{2}-b^{2})}{9.(a^{3}-b{3})}= \\
=\frac{2.\cancel{3}.\cancel{5}.\cancel{(a^{2}+a.b+b^{2})}.\cancel{(a-b)}.\cancel{(a+b)}}{3^{\cancel{2}}.\cancel{5}.\cancel{(a+b)}.\cancel{(a-b)}.\cancel{(a^{2}+a.b+b^{2})}}=\frac{2}{3}[/tex]