Trupmenos

Trupmeninių reiškinių prastinimas

1. ${\displaystyle \frac{8xy}{12c}}={\displaystyle \frac{2xy}{3x}}$

2. ${\displaystyle \frac{4a^{2}p}{3a^7{p}^3}}={\displaystyle \frac{4}{3a^5{p}^2}}$

3. ${\displaystyle \frac{4(x+y)}{5(x+y)}}={\displaystyle \frac{4}{5}}$

4. ${\displaystyle \frac{(a-2{)}^3}{a-2}}={\displaystyle \frac{(a-2{)}^2}{1}}=a^2-4a+4$

Trupmenų daugyba ir dalyba

1. ${\displaystyle \frac{3x^2}{2m}}\cdot {\displaystyle \frac{5m^3}{4x^2}}={\displaystyle \frac{15m^2}{8x^2}}$

2. ${\displaystyle \frac{x^2-9}{16x}}:{\displaystyle \frac{4x+12}{14}}$$={\displaystyle \frac{x^2-9}{16x}}\cdot {\displaystyle \frac{14}{4x+12}}$$={\displaystyle \frac{(x-3)(x+3)}{16x}}\cdot {\displaystyle \frac{14}{4(x+3)}}$$={\displaystyle \frac{14(x-3)}{64x}}={\displaystyle \frac{7(x-3)}{32x}}={\displaystyle \frac{7x-21}{32x}}$

3. ${\displaystyle \frac{5(25-4x^2)}{3x}}:{\displaystyle \frac{2x-5}{9}}$$={\displaystyle \frac{5(5-2x)(5+2x)}{3x}}\cdot {\displaystyle \frac{9}{2x-5}}$$={\displaystyle \frac{45(5-2x)(5+2x)}{3x(2x-5)}}$$={\displaystyle \frac{-45(2x-5)(5+2x)}{3x(2x-5)}}$$={\displaystyle \frac{-45(5+2x)}{3x}}$$={\displaystyle \frac{-15(5+2x)}{x}}$$={\displaystyle \frac{-30x-75}{x}}$

4. ${\displaystyle \frac{x^2-10x+25}{3x}}\cdot {\displaystyle \frac{12x^3}{5-x}}$$={\displaystyle \frac{(x-5{)}^2}{3x}}\cdot {\displaystyle \frac{12x^3}{5-x}}$$={\displaystyle \frac{(x-5{)}^2}{x}}\cdot {\displaystyle \frac{4x^3}{5-x}}$$={\displaystyle \frac{(5-x{)}^2}{x}}\cdot {\displaystyle \frac{4x^3}{5-x}}$$={\displaystyle \frac{5-x}{1}}\cdot {\displaystyle \frac{4x^2}{1}}$$=4x^2(5-x)=20x-4x^3$

Trupmenų sudėtis ir atimtis

1. ${\displaystyle \frac{3}{x-2}}+{\displaystyle \frac{5x}{x-2}}={\displaystyle \frac{3+5x}{x-2}}$

2. ${\displaystyle \frac{x-3}{5x+2}}-{\displaystyle \frac{4x+7}{5x+2}}$$={\displaystyle \frac{x-3-(4x+7)}{5x+2}}$$={\displaystyle \frac{x-3-4x-7}{5x+2}}$$={\displaystyle \frac{-3x-10}{5x+2}}$

3. ${\displaystyle \frac{4x}{x-5}}+{\displaystyle \frac{x-2}{5-x}}$$={\displaystyle \frac{4x}{x-5}}-{\displaystyle \frac{x-2}{x-5}}$$={\displaystyle \frac{4x-x+2}{x-5}}$$={\displaystyle \frac{3x-2}{x-5}}$

4. ${\displaystyle \frac{4}{2x-6}}-{\displaystyle \frac{x+2}{x^2-9}}$$={\displaystyle \frac{4}{2(x-3)}}-{\displaystyle \frac{x+2}{(x-3)(x+3)}}$$={\displaystyle \frac{2}{x-3}}-{\displaystyle \frac{x+2}{(x-3)(x+3)}}$$={\displaystyle \frac{2(x+3)}{(x-3)(x+3)}}-{\displaystyle \frac{x+2}{(x-3)(x+3)}}$$={\displaystyle \frac{2(x+3)-(x+2)}{(x-3)(x+3)}}$$={\displaystyle \frac{2x+6-x-2}{(x-3)(x+3)}}$$={\displaystyle \frac{x+4}{(x-3)(x+3)}}$$={\displaystyle \frac{x+4}{x^2-9}}$

Sudėtingesnių trupmenų prastinimas

1. Skaidymas dauginamaisiais grupavimo būdu

   1. $4xy+12y-4x-12$$=4y(x+3)-4(x+3)$$=(4y-4)(x+3)$$=4(y-1)(x+3)$

   2. $x^2-2xc+c^2-d^2$$=(x-c{)}^2-d^2$$=(x-c+d)(x-c-d)$

   3. $3a^2-3a{b}^2+a^{2}b-b^3$$=3a(a^2-b^2)+b(a^2-b^2)$$=(3a+b)(a^2-b^2)$$=(3a+b)(a+b)(a-b)$

   4. ${\displaystyle \frac{x^3-4x^2+3x-12}{x^2+3}}$$={\displaystyle \frac{x^2(x-4)+3(x-4)}{x^2+3}}$$={\displaystyle \frac{(x^2+3)(x-4)}{x^2+3}}$$=x-4$

2. Skaidant kvaratinį trinarį dauginamaisiais

   1. ${\displaystyle \frac{x^2-4x+3}{x^2-x-6}}$$={\displaystyle \frac{(x-1)(x-3)}{(x-3)(x+2)}}$$={\displaystyle \frac{x-1}{x+2}}$

   $x^2-4x+3=0$

   $D=b^2-4ac=16-4\cdot 1\cdot 3=16-12=4$

   $x_1={\displaystyle \frac{-b+\sqrt{D}}{2a}}={\displaystyle \frac{4+2}{2}}=3$

   $x_1={\displaystyle \frac{-b-\sqrt{D}}{2a}}={\displaystyle \frac{4-2}{2}}=1$

   $x^2-x-6=0$

   $D=b^2-4ac=1-4\cdot 1\cdot (-6)=1+24=25$

   $x_1={\displaystyle \frac{-b+\sqrt{D}}{2a}}={\displaystyle \frac{1+5}{2}}=3$

   $x_1={\displaystyle \frac{-b-\sqrt{D}}{2a}}={\displaystyle \frac{1-5}{2}}=-2$

   2. ${\displaystyle \frac{x^2+3x-28}{4x^2+27x-7}}$$={\displaystyle \frac{(x+7)(x-4)}{4(x-0.25)(x+7)}}$$={\displaystyle \frac{x-4}{4x-1}}$

   $x^2+3x-28=0$

   $D=b^2-4ac=9-4\cdot 1\cdot (-28)$$=9+112=121$

   $x_1={\displaystyle \frac{-b+\sqrt{D}}{2a}}$$={\displaystyle \frac{-3+11}{2}}=4$

   $x_2={\displaystyle \frac{-b-\sqrt{D}}{2a}}$$={\displaystyle \frac{-3-11}{2}}=-7$

   $4x^2+27x-7=0$

   $D=b^2-4ac={27}^2-4\cdot 4\cdot (-7)=841$ //29

   $x_1={\displaystyle \frac{-b+\sqrt{D}}{2a}}$$={\displaystyle \frac{-27+29}{2\cdot 4}}$$=0.25$

   $x_2={\displaystyle \frac{-b-\sqrt{D}}{2a}}$$={\displaystyle \frac{-27-29}{2\cdot 4}}$$=-7$

3. ${\displaystyle \frac{y^2-2\sqrt{6}y-18}{0.5y^2-2\sqrt{6}y+9}}$$={\displaystyle \frac{(y-3\sqrt{6})(y+\sqrt{6})}{(y-3\sqrt{6})(y-\sqrt{6})}}$$={\displaystyle \frac{y+\sqrt{6}}{y-\sqrt{6}}}$

   $y^2-2\sqrt{6}y-18=0$

   $D=b^2-4ac=(2\sqrt{6}{)}^2+4\cdot 1\cdot 18=24+72=96$

   $y_1={\displaystyle \frac{-b+\sqrt{D}}{2a}}$$={\displaystyle \frac{2\sqrt{6}+4\sqrt{6}}{2}}$$={\displaystyle \frac{6\sqrt{6}}{2}}$$=3\sqrt{6}$

   $y^2-4\sqrt{6}y+18=0$

   $D=b^2-4ac=(4\sqrt{6}{)}^2-4\cdot 1\cdot 18=16\cdot 6-72=24$

   $y_1={\displaystyle \frac{-b+\sqrt{D}}{2a}}$$={\displaystyle \frac{4\sqrt{6}+2\sqrt{6}}{2}}$$={\displaystyle \frac{6\sqrt{6}}{2}}$$=3\sqrt{6}$

   $y_2={\displaystyle \frac{-b+\sqrt{D}}{2a}}$$={\displaystyle \frac{4\sqrt{6}-2\sqrt{6}}{2}}$$={\displaystyle \frac{2\sqrt{6}}{2}}$$=\sqrt{6}$