Trupmeninės lygtys

${\displaystyle \frac{a}{b}}=0$

$\{\begin{array}{l}a=0\\ b\ne 0\end{array}$

Paprastos trupmeninės lygtys

1. ${\displaystyle \frac{x^2-x-6}{x-3}}=0$

   $\{\begin{array}{l}{x}^2-x-6=0\\ x-3\ne 0\end{array}$

   $x^2-x-6=0$

   $D=1-4\cdot 1\cdot (-6)=1+24=25$

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

   $x_2={\displaystyle \frac{-b-\sqrt{D}}{2a}}={\displaystyle \frac{1-5}{2\cdot 1}}=-2$

   $x-3\ne 0$

   $x\ne 3$

   Ats.: $x=-2$

   
   

2. ${\displaystyle \frac{x-2}{x+5}}=0$

   $\{\begin{array}{l}x-2=0\\ x+5\ne 0\end{array}$

   $x-2=0$

   $x=2$

   $x+5\ne 0$

   $x\ne -5$

   Ats.: $x=2$

   
   

3. ${\displaystyle \frac{2}{x}}={\displaystyle \frac{x}{3}}$

   ${\displaystyle \frac{2\cdot 3}{x\cdot 3}}-{\displaystyle \frac{x\cdot x}{3\cdot x}}=0$

   ${\displaystyle \frac{6-x^2}{3x}}=0$

   $\{\begin{array}{l}6-x^2=0\\ 3x\ne 0\end{array}$

   $6-x^2=0$

   $x^2=6$

   $x_1=\sqrt{6}$

   $x_2=-\sqrt{6}$

   $3x\ne 0$

   $x\ne 0$

   Ats.: $x=\sqrt{6}$, $x=-\sqrt{6}$

Sudėtingesnės trupmeninės lygtys

Sprendimo žingniai:

1. Sukeliame visus narius į kairę pusę, kad dešinėje būtų 0.

2. Dauginame iš bendrojo vardiklio.

3. Sudarome sistemą

1. ${\displaystyle \frac{3x}{3-x}}+{\displaystyle \frac{9}{x-3}}=x$

   ${\displaystyle \frac{3x}{3-x}}-{\displaystyle \frac{9}{3-x}}=x$

   ${\displaystyle \frac{3x-9}{3-x}}=x$

   ${\displaystyle \frac{3(x-3)}{3-x}}=x$

   $-{\displaystyle \frac{3(3-x)}{3-x}}=x$

   $-3=x$

   Ats.: $x=-3$

   
   

2. ${\displaystyle \frac{4}{x^2-2x}}-{\displaystyle \frac{2}{x-2}}+{\displaystyle \frac{1}{2}}=0$

   ${\displaystyle \frac{4}{x(x-2)}}-{\displaystyle \frac{2}{x-2}}+{\displaystyle \frac{0.5}{1}}=0$

   ${\displaystyle \frac{4}{x(x-2)}}-{\displaystyle \frac{2(x)}{x(x-2)}}+{\displaystyle \frac{0.5x(x-2)}{x(x-2)}}=0$

   ${\displaystyle \frac{4-2x+0.5x(x-2)}{x(x-2)}}=0$

   ${\displaystyle \frac{4-2x+0.5x^2-x}{x(x-2)}}=0$

   ${\displaystyle \frac{0.5x^2-3x+4}{x(x-2)}}=0$

   $\{\begin{array}{l}0.5x^2-3x+4=0\\ x(x-2)\ne 0\end{array}$

   $0.5x^2-3x+4=0$

   $D=9-4\cdot 0.5\cdot 4=1$

   $x_1={\displaystyle \frac{3+1}{1}}=4$

   $x_2={\displaystyle \frac{3-1}{1}}=2$

   $x(x-2)\ne 0$

   $x\ne 0$

   $x\ne 2$

   $\{\begin{array}{l}x=4,x=2\\ x\ne 0,x\ne 2\end{array}$

   Ats.: $x=4$

   
   

3. ${\displaystyle \frac{x^2-6x+8}{2x(x-2)}}=0$

   $\{\begin{array}{l}{x}^2-6x+8=0\\ 2x(x-2)\ne 0\end{array}$

   $x^2-6x+8=0$

   $D=36-4\cdot 1\cdot 8=36-32=4$

   $x_1={\displaystyle \frac{6+2}{2\cdot 1}}=4$

   $x_2={\displaystyle \frac{6-2}{2\cdot 1}}=2$

   $2x(x-2)\ne 0$

   $x(x-2)\ne 0$

   $x\ne 0$

   $x\ne 2$

   $\{\begin{array}{l}x=4,x=2\\ x\ne 0,x\ne 2\end{array}$

   Ats.: $x=4$

   
   

4. ${\displaystyle \frac{2y+1}{y-1}}+{\displaystyle \frac{y+1}{2y+1}}={\displaystyle \frac{5y+4}{2y^2-y-1}}$

   $2y^2-y-1=0$

   $D=1+4\cdot 2\cdot 1=9$

   $y_1={\displaystyle \frac{1+3}{2\cdot 2}}=1$

   $y_2={\displaystyle \frac{1-3}{2\cdot 2}}=-0.5$

   ${\displaystyle \frac{2y+1}{y-1}}+{\displaystyle \frac{y+1}{2y+1}}={\displaystyle \frac{5y+4}{2(y-1)(y+0.5)}}$

   ${\displaystyle \frac{2y+1}{y-1}}+{\displaystyle \frac{y+1}{2y+1}}-{\displaystyle \frac{5y+4}{2(y-1)(y+0.5)}}=0$

   ${\displaystyle \frac{2y+1}{y-1}}+{\displaystyle \frac{y+1}{2y+1}}-{\displaystyle \frac{5y+4}{(y-1)(2y+1)}}=0$

   ${\displaystyle \frac{(2y+1)(2y+1)+(y+1)(y-1)-(5y+4)}{(y-1)(2y+1)}}=0$

   ${\displaystyle \frac{4y^2+2y+2y+1+y^2-1-5y-4}{(y-1)(2y+1)}}=0$

   ${\displaystyle \frac{5y^2-y-4}{(y-1)(2y+1)}}=0$

   $\{\begin{array}{l}5y^2-y-4=0\\ (y-1)(2y+1)\ne 0\end{array}$

   $5y^2-y-4=0$

   $D=1+4\cdot 5\cdot 4=81$

   $y_1={\displaystyle \frac{1+9}{2\cdot 5}}=1$

   $y_2={\displaystyle \frac{1-9}{2\cdot 5}}=-0.8$

   $(y-1)(2y+1)\ne 0$

   $y\ne 1$

   $y\ne -{\displaystyle \frac{1}{2}}$

   $\{\begin{array}{l}y=1,y=-0.8\\ y\ne 1,y\ne -0.5\end{array}$

   Ats.: $y=-0.8$

Dar vienas trupmeninių lygčių sprendimo būdas

1. Trupmenų vardiklius iškaidome daugikliais

2. Dauginame lygtį iš visų vardikliuose esančių reiškinių sandaugos

3. Patikriname gautus sprendinius (ar nėra dalybos iš 0)

1. ${\displaystyle \frac{1}{2}}-{\displaystyle \frac{2}{x-2}}={\displaystyle \frac{4}{2x-x^2}}$

   ${\displaystyle \frac{1}{2}}-{\displaystyle \frac{2}{x-2}}={\displaystyle \frac{4}{x(2-x)}}$

   ${\displaystyle \frac{1}{2}}+{\displaystyle \frac{2}{2-x}}={\displaystyle \frac{4}{x(2-x)}}$     $/\cdot 2x(2-x)$

   $2x(2-x)\ne 0$

   $x\ne 0$

   $x\ne 2$

   $x(2-x)+4x=8$

   $2x-x^2+4x=8$

   $6x-x^2=8$

   $x^2-6x+8=0$

   $D=6^2-4\cdot 1\cdot 8=4$

   $x_1={\displaystyle \frac{6+2}{2}}=4$

   $x_2={\displaystyle \frac{6-2}{2}}=2$

   Ats.: $x=4$

2. ${\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{9}{2x+8}}+{\displaystyle \frac{x+1}{2-2x}}=0$

   ${\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{9}{2(x+4)}}+{\displaystyle \frac{x+1}{2(1-x)}}=0$

   ${\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{9}{2(x+4)}}-{\displaystyle \frac{x+1}{2(x-1)}}=0$     $/\cdot 2(x-1)(x+4)$

   $2(x-1)(x+4)\ne 0$

   $x\ne 1$

   $x\ne -4$

   $2(x+4)+9(x-1)-(x+1)(x+4)=0$

   $2x+8+9x-9-x^2-5x-4=0$

   $-x^2+6x-5=0$

   $x^2-6x+5=0$

   $D=6^2-4\cdot 1\cdot 5=36-20=16$

   $x_1={\displaystyle \frac{6+4}{2}}=5$

   $x_2={\displaystyle \frac{6-4}{2}}=1$

   Ats.: $x=5$

Tekstiniai uždaviniai

1. Kateris, nuplaukdamas 36 km upe pasroviui, užtrunka tiek pat laiko, kiek nuplaukdamas 20 km prieš srovę. Apskaičiuokite katerio savąjį greitį, jei upės tėkmės greitis lygus 2 km/h. (Tempus Matamtika 10, I dalis, 183psl.)

   	$s,km$	$v,km/h$	$t,h$
   Pasroviui	36	$v+2$	${\displaystyle \frac{36}{x+2}}$
   Prieš srovę	20	$v-2$	${\displaystyle \frac{20}{x-2}}$

   Katerio savasis greitis lygus $xkm/h$

   ${\displaystyle \frac{36}{x+2}}={\displaystyle \frac{20}{x-2}}$     $/\cdot (x+2)(x-2)$

   $(x+2)(x-2)\ne 0$

   $x\ne 2$

   $x\ne -2$

   $36(x-2)=20(x+2)$

   $36x-72=20x+40$

   $36x-20x=40+72$

   $16x=112$

   $x={\displaystyle \frac{112}{16}}$

   $x=7$

   Ats.: $7km/h$

2. Kateris nuplaukė 18 km pasroviui ir 6 km prieš srovę. Kelionė truko 4 valan- das. Koks buvo katerio savasis greitis, jei upės tėkmės greitis lygus 3 km/h? (Tempus Matamtika 10, I dalis, 183psl.)

   	$s,km$	$v,km/h$	$t,h$
   Pasroviui	18	$v+3$	${\displaystyle \frac{18}{x+3}}$
   Prieš srovę	6	$v-3$	${\displaystyle \frac{6}{x-3}}$

   Katerio savasis greitis lygus $xkm/h$

   ${\displaystyle \frac{18}{x+3}}={\displaystyle \frac{6}{x-3}}$     $/\cdot (x+3)(x-3)$

   $(x+3)(x-3)\ne 0$

   $x\ne 3$

   $x\ne -3$

   $18(x-3)=6(x+3)$

   $18x-54=6x+18$

   $18x-6x=18+54$

   $12x=72$

   $x={\displaystyle \frac{72}{12}}$

   $x=6$

   Ats.: $6km/h$

3. 21 km pasroviui kateris plaukia 15 min trumpiau negu tą patį atstumą prieš srovę. Apskaičiuokite katerio savąjį greitį, jei upės tėkmės greitis lygus 1 km/h. (Tempus Matamtika 10, I dalis, 183psl.)

   	$s,km$	$v,km/h$	$t,h$
   Pasroviui	21	$v+1$	${\displaystyle \frac{21}{x+1}}$
   Prieš srovę	21	$v-1$	${\displaystyle \frac{21}{x-1}}$

   Katerio savasis greitis lygus $xkm/h$

   ${\displaystyle \frac{21}{x+1}}+{\displaystyle \frac{1}{4}}={\displaystyle \frac{21}{x-1}}$     $/\cdot 4(x+1)(x-1)$

   $4(x+1)(x-1)\ne 0$

   $x\ne 1$

   $x\ne -1$

   $21\cdot 4(x-1)+(x+1)(x-1)=21\cdot 4(x+1)$

   $84x-84+x^2-1=84x+84$

   $x^2-1=168$

   $x^2=169$

   $x=\sqrt{169}$

   $x=13$

   Ats.: $13km/h$

Bendro darbo uždaviniai

• Uždaviniuose dažniausiai yra kalba a apie darbą / baseinų pripildymą

• Darbas konkrečiai neįvardintas („turi atlikti tam tikrą darbą“, „turi suarti lauką“)

• Visą darbą laikome vienetu

Šiuos uždavinius patogu spręsti naudojant lentelę

1. Įmonėje du kopijavimo aparatai, veikdami vienu metu, gali per 10min atspausdinti reikiamą skaičių kopijų. Per kiek minučių tiek pat kopijų atspausdintų kiekvienas kopijavimo aparatas, veikdamas atskirai, jei pirmasis aparatas visas kopijas atspausdina 15 min anksčiau negu antrasis?

   	Laikas, atliekant visą darbą atskirai, $min$	Per $1min$ atliekama darbo dalis	Per $10min$ atliekama darbo dalis
   Pirmasis	$x-15$	${\displaystyle \frac{1}{x-15}}$	${\displaystyle \frac{10}{x-15}}$
   Antrasis	$x$	${\displaystyle \frac{1}{x}}$	${\displaystyle \frac{10}{x}}$

   ${\displaystyle \frac{10}{x-15}}+{\displaystyle \frac{10}{x}}=1$     $/\cdot x(x-15)$

   $10x+10(x-15)=x(x-15)$

   $10x+10x-150=x^2-15x$

   $20x-150=x^2-15x$

   $0=x^2-15x-20x+150$

   $0=x^2-35x+150$

   $x^2-35x+150=0$

   $D={35}^2-4\cdot 1\cdot 150=1225-600=625$

   $x_1={\displaystyle \frac{35+25}{2}}=30$

   $x_2={\displaystyle \frac{35-25}{2}}=5$

   Ats.: $x=30$

Lygčių sistemos, kai 1 lygtis - trupmeninė

1. $\{\begin{array}{l}x-y=1\\ {\displaystyle \frac{1}{x+2}}-y=3\end{array}$

   $\{\begin{array}{l}x=1+y\\ {\displaystyle \frac{1}{x+2}}-y=3\end{array}$

   ${\displaystyle \frac{1}{1+y+2}}-y=3$

   ${\displaystyle \frac{1}{y+3}}-{\displaystyle \frac{y}{1}}-{\displaystyle \frac{3}{1}}=0$     $/\cdot y+3$

   $y+3\ne 0$

   $y\ne -3$

   $y\ne 0$

   $1-y(y+3)-3(y+3)=0$

   $1-y^2-3y-3y-9=0$

   $1-y^2-6y-9=0$

   $-y^2-6y-8=0$

   $y^2+6y+8=0$

   $D=6^2-4\cdot 1\cdot 8=36-32=4$

   $y_1={\displaystyle \frac{-6+2}{2}}=-2$     $x_1=1-2=-1$

   $y_2={\displaystyle \frac{-6-2}{2}}=-4$     $x_2=1-4=-3$

   Ats.: $(-1;-2),(-3;-4)$

2. $\{\begin{array}{l}x-y=1\\ {\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}={\displaystyle \frac{3}{2}}\end{array}$

   $\{\begin{array}{l}x=1+y\\ {\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}-{\displaystyle \frac{3}{2}}=0\end{array}$

   ${\displaystyle \frac{1}{1+y}}+{\displaystyle \frac{1}{y}}-{\displaystyle \frac{3}{2}}=0$     $/\cdot 2y(1+y)$

   $2y(1+y)\ne 0$

   $y\ne 0$

   $y\ne -1$

   $2y+2(1+y)-3y(1+y)=0$

   $2y+2+2y-3y-3y^2=0$

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

   $-3y^2+y+2=0$

   $3y^2-y-2=0$

   $D=1+24=25$

   $y_1={\displaystyle \frac{1+5}{6}}=1$     $x_1=1+1=2$

   $y_2={\displaystyle \frac{1-5}{6}}=-{\displaystyle \frac{2}{3}}$     $x_2=1-{\displaystyle \frac{2}{3}}={\displaystyle \frac{1}{3}}$

   Ats.: $(2;1),({\displaystyle \frac{1}{3}};-{\displaystyle \frac{2}{3}})$