Trupmenos

Trupmeninių reiškinių prastinimas

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

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

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

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

Trupmenų daugyba ir dalyba

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

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

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

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

Trupmenų sudėtis ir atimtis

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

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

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

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

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

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

2. Skaidant kvaratinį trinarį dauginamaisiais

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

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

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

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

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

   $x^2-x-6=0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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