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Trys dviratininkai kas valandą išvažiuoja iš tos pačios vietos ir važiuoja viena kryptimi.
Pirmojo dviratininko greitis 12 km/h, antrojo – 10 km/h. Trečiasis dviratininkas, važiuodamas
greičiau nei pirmasis, pirmiausia pavijo antrąjį, o praėjus dar 2 valandoms – pirmąjį
dviratininką. Koks trečiojo dviratininko greitis?
Sprendimas.
x - trečiojo dviratininko greitis.
Tarkim trečiasis dviratininkas pavijo antrąjį po t valandų, kai išvažiavo pats. Tuo momentu antrasis dviratininkas, važiavęs greičiu 10 km/h, buvo užtrukęs 1 + t valandų, kadangi buvo išvažiavęs valanda anksčiau.
xt = 10 (t + 1).
Dar po dviejų valandų pavijo pirmąjį, važiavusį 12 km/h greičiu. Trečiojo sugaištas laikas: t + 2, o pirmojo t + 4, kadangi išvažiavo dviem valandom anksčiau.
x(t+2) = 12 (t + 4).
Dviejų lygčių sistema:
xt = 10 (t + 1)
x(t+2) = 12 (t + 4).
Iš pirmos lygties išsireiškiam t:
x* t =
10* (t+1)
10* (t+1)
|
x* t = 10* (t+1)$$x\cdot t$$ = $$10\cdot (t+1)$$
x* t = 10* t+ 10* 1$$x\cdot t$$ = $$10\cdot t+10\cdot 1$$
$${\normalsize 10\cdot 1}$$ = $${\normalsize 10}$$
x* t = 10* t+10$$x\cdot t$$ = $$10\cdot t+10$$
x* t- 10* t = 10$$x\cdot t-10\cdot t$$ = $$10$$
$${\normalsize x\cdot t-10\cdot t}$$ = $${\normalsize t\cdot (x-10)}$$ Paaiškinimas:
Paaiškinimas: t iškeltas prieš skliaustus
t* (x-10) = 10$$t\cdot (x-10)$$ = $$10$$
t =
$$t$$ = $$\frac{10}{x-10}$$
| 10 |
| / (x-10) |
$$x\cdot t$$ = $$10\cdot (t+1)$$
$$x\cdot t$$ = $$10\cdot t+10$$
$$x\cdot t-10\cdot t$$ = $$10$$
$$t\cdot (x-10)$$ = $$10$$
$$t$$ = $$\frac{10}{x-10}$$
Gautą t išraišką statome į antrą lygtį:
x* (t+2) =
12* (t+4)
12* (t+4)
|
|
x* (t+2) = 12* (t+4)$$x\cdot (t+2)$$ = $$12\cdot (t+4)$$
x* (t+2) = 12* t+ 12* 4$$x\cdot (t+2)$$ = $$12\cdot t+12\cdot 4$$
$${\normalsize 12\cdot 4}$$ = $${\normalsize 48}$$
x* (t+2) = 12* t+48$$x\cdot (t+2)$$ = $$12\cdot t+48$$
x* t+ x* 2 = 12* t+48$$x\cdot t+x\cdot 2$$ = $$12\cdot t+48$$
$${\normalsize x\cdot 2}$$ = $${\normalsize 2\cdot x}$$
x* t+ 2* x = 12* t+48$$x\cdot t+2\cdot x$$ = $$12\cdot t+48$$
Paaiškinimas: Keitimas $${\normalsize t}$$ = $${\normalsize \frac{10}{x-10}}$$.
| x* 10 |
| / (x-10) |
| 12* 10 |
| / (x-10) |
$${\normalsize 12\cdot 10}$$ = $${\normalsize 120}$$
| x* 10 |
| / (x-10) |
| 120 |
| / (x-10) |
| x* 10 |
| / (x-10) |
| 120 |
| / (x-10) |
| x* 10 |
| / (x-10) |
| 120 |
| / (x-10) |
$${\normalsize \frac{x\cdot 10}{x-10}-\frac{120}{x-10}}$$ = $${\normalsize \frac{10\cdot x-120}{x-10}}$$
| ( 10* x-120) |
| / (x-10) |
10* x-120 = (x-10)* (48- 2* x)$$10\cdot x-120$$ = $$(x-10)\cdot (48-2\cdot x)$$
10* x-120 = x* (48- 2* x)- 10* (48- 2* x)$$10\cdot x-120$$ = $$x\cdot (48-2\cdot x)-10\cdot (48-2\cdot x)$$
10* x-120 = x* 48- x* 2* x- 10* (48- 2* x)$$10\cdot x-120$$ = $$x\cdot 48-x\cdot 2\cdot x-10\cdot (48-2\cdot x)$$
10* x-120 = x* 48- 2* x^2- 10* (48- 2* x)$$10\cdot x-120$$ = $$x\cdot 48-2\cdot x^{2}-10\cdot (48-2\cdot x)$$
10* x-120 = 48* x- 2* x^2- 10* (48- 2* x)$$10\cdot x-120$$ = $$48\cdot x-2\cdot x^{2}-10\cdot (48-2\cdot x)$$
10* x-120 = 48* x- 2* x^2- 10* 48+ 10* 2* x$$10\cdot x-120$$ = $$48\cdot x-2\cdot x^{2}-10\cdot 48+10\cdot 2\cdot x$$
$${\normalsize 10\cdot 48}$$ = $${\normalsize 480}$$
10* x-120 = 48* x- 2* x^2-480+ 10* 2* x$$10\cdot x-120$$ = $$48\cdot x-2\cdot x^{2}-480+10\cdot 2\cdot x$$
10* x-120 = 48* x- 2* x^2-480+ 20* x$$10\cdot x-120$$ = $$48\cdot x-2\cdot x^{2}-480+20\cdot x$$
2* x^2+ 10* x-120 = 48* x-480+ 20* x$$2\cdot x^{2}+10\cdot x-120$$ = $$48\cdot x-480+20\cdot x$$
2* x^2- 20* x+ 10* x-120 = 48* x-480$$2\cdot x^{2}-20\cdot x+10\cdot x-120$$ = $$48\cdot x-480$$
2* x^2- 20* x+ 10* x- 48* x-120 = -480$$2\cdot x^{2}-20\cdot x+10\cdot x-48\cdot x-120$$ = $$-480$$
2* x^2- 20* x+ 10* x- 48* x-120+480 = 0$$2\cdot x^{2}-20\cdot x+10\cdot x-48\cdot x-120+480$$ = $$0$$
$${\normalsize -20\cdot x+10\cdot x}$$ = $${\normalsize -10\cdot x}$$
2* x^2- 10* x- 48* x-120+480 = 0$$2\cdot x^{2}-10\cdot x-48\cdot x-120+480$$ = $$0$$
$${\normalsize -10\cdot x-48\cdot x}$$ = $${\normalsize -58\cdot x}$$
2* x^2- 58* x-120+480 = 0$$2\cdot x^{2}-58\cdot x-120+480$$ = $$0$$
$${\normalsize -120+480}$$ = $${\normalsize 360}$$
2* x^2- 58* x+360 = 0$$2\cdot x^{2}-58\cdot x+360$$ = $$0$$
$${\normalsize 2\cdot x^{2}-58\cdot x+360}$$ = $${\normalsize 2\cdot (x-20)\cdot (x-9)}$$ Paaiškinimas:
Paaiškinimas: 2 iškeltas prieš skliaustus, skliaustuose liko $${\normalsize x^{2}-29\cdot x+180}$$.
Kvadratinis trinaris $${\normalsize a\cdot x^{2}+b\cdot x+c}$$, kur
a = 1, b = -29, c = 180.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 841-720}$$ = 121.
x1 = $${\normalsize \frac{29+\sqrt {121}}{2\cdot 1} = \frac{29+11}{2} = \frac{40}{2}}$$ = 20
x2 = $${\normalsize \frac{29-\sqrt {121}}{2\cdot 1} = \frac{29-11}{2} = \frac{18}{2}}$$ = 9
Kvadratinis trinaris $${\normalsize a\cdot x^{2}+b\cdot x+c}$$, kur
a = 1, b = -29, c = 180.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 841-720}$$ = 121.
x1 = $${\normalsize \frac{29+\sqrt {121}}{2\cdot 1} = \frac{29+11}{2} = \frac{40}{2}}$$ = 20
x2 = $${\normalsize \frac{29-\sqrt {121}}{2\cdot 1} = \frac{29-11}{2} = \frac{18}{2}}$$ = 9
2* (x-20)* (x-9) = 0$$2\cdot (x-20)\cdot (x-9)$$ = $$0$$
$$x\cdot (t+2)$$ = $$12\cdot (t+4)$$
$$x\cdot t+2\cdot x$$ = $$12\cdot t+48$$
$$\frac{x\cdot 10}{x-10}+2\cdot x$$ = $$\frac{12\cdot 10}{x-10}+48$$
$$\frac{x\cdot 10}{x-10}-\frac{120}{x-10}$$ = $$48-2\cdot x$$
$$\frac{10\cdot x-120}{x-10}$$ = $$(48-2\cdot x)$$
$$10\cdot x-120$$ = $$(x-10)\cdot (48-2\cdot x)$$
$$10\cdot x-120$$ = $$48\cdot x-2\cdot x^{2}-480+20\cdot x$$
$$2\cdot x^{2}-20\cdot x+10\cdot x-48\cdot x-120+480$$ = $$0$$
$$2\cdot x^{2}-58\cdot x-120+480$$ = $$0$$
$$2\cdot x^{2}-58\cdot x+360$$ = $$0$$
$$2\cdot (x-20)\cdot (x-9)$$ = $$0$$
$$x$$ = $$20$$
$$x$$ = $$9$$
Gavome x = 9 ir x = 20.
Kadangi trečiojo dviratininko greitis didžiausias iš visų, tinka tik x = 20
Atsakymas: 20 km/h