1)      Vypočti:

\displaystyle a)\quad \frac{9a}{4b}:\left( -\frac{3a}{b} \right)=

 

\displaystyle b)\quad \frac{2rs}{5p}:\frac{3rp}{2s}=

 

\displaystyle c)\quad \frac{8x}{21{{y}^{2}}}:\frac{6{{x}^{2}}}{7y}=

 

\displaystyle d)\quad -8{{a}^{2}}{{b}^{4}}:\left( -\frac{4{{b}^{3}}}{3a} \right)=
\displaystyle e)\quad \left( a\cdot \frac{1}{b} \right):\left( b\cdot \frac{1}{a} \right)=

 

\displaystyle f)\quad \frac{r}{x}:\frac{2r}{{{x}^{2}}}+r\cdot \frac{3}{x}=

 

\displaystyle g)\quad \frac{12uv}{25z}:8{{z}^{2}}=

 

\displaystyle h)\quad \frac{18{{a}^{2}}b}{5cd}:\frac{6ab}{5c{{d}^{3}}}=
\displaystyle i)\quad -\frac{1}{{{a}^{2}}{{b}^{3}}c}:\left( -\frac{1}{{{a}^{3}}{{b}^{2}}c} \right)=

 

\displaystyle j)\quad -\frac{21u{{v}^{2}}}{2xz}:14{{u}^{2}}vx=

 

\displaystyle k)\quad \left( \frac{2ab}{m}\cdot \frac{{{a}^{2}}m}{3bn} \right):\frac{2a}{3n}=

 

\displaystyle l)\quad \frac{5{{r}^{3}}}{4s}:\left( 3{{r}^{2}}\cdot \frac{5r}{6s} \right)=

 

2)      Vypočti:

\displaystyle a)\quad \frac{x\left( a+b \right)}{2a}:\frac{{{x}^{2}}}{a}=

 

\displaystyle b)\quad \frac{2c-2}{{{d}^{2}}}:\frac{c-1}{d}=

 

\displaystyle c)\quad \frac{{{t}^{2}}-2t}{3}:\frac{t}{6}=

 

\displaystyle d)\quad \frac{3r}{s+5}:\frac{r}{s-2}=

 

\displaystyle e)\quad \frac{{{x}^{2}}-xy}{y}:\frac{x-y}{xy}=

 

\displaystyle f)\quad \frac{a+b}{a-b}:\frac{b+a}{b-a}=
\displaystyle g)\quad \frac{b-2}{a+b}:\frac{3b-6}{2a+2b}=

 

\displaystyle h)\quad \frac{{{x}^{2}}y}{5\left( x+1 \right)}:\frac{x{{y}^{2}}}{2x+2}=

 

\displaystyle i)\quad \frac{{{y}^{2}}+y}{4y-12}:\frac{7\left( y+1 \right)}{4y}=

 

\displaystyle j)\quad \frac{{{a}^{2}}+3}{2a}:\frac{{{a}^{3}}+3a}{4{{a}^{2}}}=

 

\displaystyle k)\quad \frac{2m+6}{{{m}^{2}}}:\frac{m+3}{{{m}^{2}}-mn}=

 

\displaystyle l)\quad \frac{15+15r}{{{r}^{2}}-1}:\frac{3r+3}{{{r}^{3}}-r}=
\displaystyle m)\quad \frac{{{\left( x+y \right)}^{2}}}{{{x}^{2}}-{{y}^{2}}}:\frac{x+y}{x-y}=

 

\displaystyle n)\quad \frac{{{v}^{2}}-1}{{{v}^{3}}}:\frac{{{\left( v+1 \right)}^{2}}}{{{v}^{2}}}=

 

\displaystyle o)\quad \frac{r+3}{r-3}:\frac{{{r}^{2}}+3r}{2{{r}^{2}}-18}=

 

\displaystyle p)\quad \frac{2x+2y}{3y-6}:\frac{x+y}{y-2}=

 

\displaystyle q)\quad \frac{{{a}^{2}}+ab}{a}:\frac{b}{ab+{{b}^{2}}}=

 

\displaystyle r)\quad \frac{2\left( a+b \right)}{3a-3b}:\frac{6a+6b}{{{a}^{2}}-ab}=

 

3)      Vypočti:

\displaystyle a)\quad \frac{5-5x}{{{\left( 1+x \right)}^{2}}}:\frac{10\left( 1-{{x}^{2}} \right)}{3\left( 1+x \right)}=

 

\displaystyle b)\quad \frac{{{u}^{2}}-{{v}^{2}}}{{{\left( u+v \right)}^{2}}}:\frac{5u-5v}{4u+4v}=

 

\displaystyle c)\quad \frac{{{a}^{2}}+2ab+{{b}^{2}}}{{{a}^{2}}+6a+9}:\frac{{{a}^{2}}-{{b}^{2}}}{\left( a+3 \right)\left( a-b \right)}=

 

\displaystyle d)\quad \frac{p+q}{p-q}:\frac{{{p}^{2}}-{{q}^{2}}}{{{p}^{2}}-2pq+{{q}^{2}}}=

 

\displaystyle e)\quad \frac{3{{a}^{2}}+12a+12}{a-2}:\frac{6\left( a+2 \right)}{{{a}^{2}}-4}=

 

\displaystyle f)\quad \frac{a\left( {{x}^{2}}-{{y}^{2}} \right)}{{{\left( x+y \right)}^{2}}}:\frac{a{{\left( x-y \right)}^{2}}}{3\left( x+y \right)}=
\displaystyle g)\quad \frac{v-3}{{{v}^{2}}+v}:\frac{3v-9}{v\left( 1+v \right)}=

 

\displaystyle h)\quad \frac{{{a}^{2}}-25}{{{a}^{2}}+10a+25}:\frac{7a-35}{{{a}^{2}}+5a}=

 

\displaystyle i)\quad \frac{x+2}{4x}:\frac{{{x}^{2}}}{x-2}=

 

\displaystyle j)\quad \frac{{{x}^{2}}-4{{y}^{2}}}{{{x}^{2}}-xy}:\frac{{{x}^{2}}+2xy}{x-y}=

 

\displaystyle k)\quad \left( \frac{u}{v}-\frac{v}{u} \right):\frac{u+v}{uv}=

 

\displaystyle l)\quad \left( \frac{1}{b}-\frac{1}{a} \right):\left( {{a}^{2}}-ab \right)=

 

4)      Vypočti:

\displaystyle a)\quad \left( \frac{x}{2}-\frac{2}{x} \right):\frac{2+x}{2x}=

 

\displaystyle b)\quad \left( y+2 \right):\left( \frac{1}{y}+\frac{1}{2} \right)=

 

\displaystyle c)\quad \left( \frac{1}{n}-\frac{1}{m} \right):\frac{3m-3n}{{{m}^{2}}}=
\displaystyle d)\quad \left( 3+\frac{3s}{r} \right):\left( \frac{s}{{{r}^{2}}}+\frac{1}{r} \right)=

 

\displaystyle e)\quad \left( 3-\frac{1}{x} \right):\frac{9{{x}^{2}}-1}{{{x}^{3}}}=

 

\displaystyle f)\quad \left( {{p}^{2}}-\frac{1}{{{p}^{2}}} \right):\left( p+\frac{1}{p} \right)=

 

5)      Vypočti:

\displaystyle a)\quad \left( \frac{3x-2}{2x+3}+\frac{2x+1}{2x+3} \right):\frac{5x-1}{3+2x}=

 

\displaystyle b)\quad \left( \frac{{{a}^{2}}{{b}^{2}}}{a+b}:\frac{ab}{2a+2b} \right)\cdot \left( \frac{a}{b}+\frac{b}{a} \right)=

 

\displaystyle c)\quad \left( \frac{a}{x-a}-\frac{a}{x+a} \right):\frac{2{{a}^{2}}}{{{x}^{2}}+2ax+{{a}^{2}}}=

 

\displaystyle d)\quad \left( \frac{1}{1-a}-1 \right):\left( a-\frac{1-2{{a}^{2}}}{1-a}+1 \right)=
\displaystyle e)\quad \left( \frac{a+1}{2a-2}+\frac{6}{2{{a}^{2}}-2}-\frac{a+3}{2a+2} \right):\frac{3}{4{{a}^{2}}-4}=

 

\displaystyle f)\quad \left( \frac{b}{{{a}^{2}}+ab}-\frac{2}{a+b}+\frac{a}{{{b}^{2}}+ab} \right):\left( \frac{b}{a}-2+\frac{a}{b} \right)=

 

\displaystyle g)\quad \left( m+1-\frac{1}{1-m} \right):\left( m-\frac{{{m}^{2}}}{m-1} \right)=

 

\displaystyle h)\quad \left( \frac{{{c}^{2}}+{{d}^{2}}}{c}-2d \right):\left[ \left( \frac{1}{{{d}^{2}}}-\frac{1}{{{c}^{2}}} \right)\cdot \frac{cd}{c+d} \right]=
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