15 uždavinys
Išspręskite lygtį ir nelygybę.
1. $$4^{x}-3\cdot 2^{x}-4 = 0$$.
Sprendimas:
4^x- 3* 2^x-4 =
0
0
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4^x- 3* 2^x-4 = 0$$4^{x}-3\cdot 2^{x}-4$$ = $$0$$
( 2^x)^2- 3* 2^x-4 = 0$$(2^{x})^{2}-3\cdot 2^{x}-4$$ = $$0$$
Paaiškinimas: ( 2^x-4)* ( 2^x+1) = 0$$(2^{x}-4)\cdot (2^{x}+1)$$ = $$0$$
2^x-4 = 0$$2^{x}-4$$ = $$0$$
2^x = 0+4$$2^{x}$$ = $$0+4$$
2^x = 4$$2^{x}$$ = $$4$$
log(2, 2^x) = log(2,4)$$log_{2}(2^{x})$$ = $$log_{2}(4)$$
x = log(2,4)$$x$$ = $$log_{2}(4)$$
x = 2$$x$$ = $$2$$
2^x+1 = 0$$2^{x}+1$$ = $$0$$
2^x = 0-1$$2^{x}$$ = $$0-1$$
Šaknų nėra
2^x = -1$$2^{x}$$ = $$-1$$
Atsakymas: x = 2
2. $$lg(x-1)+lg(x) < lg(20)$$.
Sprendimas:
lg(x-1)+lg(x) <
lg(20)
lg(20)
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lg(x-1)+lg(x) < lg(20)$$lg(x-1)+lg(x)$$ < $$lg(20)$$
Paaiškinimas: lg( (x-1)* x) < lg(20)$$lg((x-1)\cdot x)$$ < $$lg(20)$$
lg( x* x- 1* x) < lg(20)$$lg(x\cdot x-1\cdot x)$$ < $$lg(20)$$
lg( x^2- 1* x) < lg(20)$$lg(x^{2}-1\cdot x)$$ < $$lg(20)$$
lg( x^2-x) < lg(20)$$lg(x^{2}-x)$$ < $$lg(20)$$
Paaiškinimas: x^2-x < 20$$x^{2}-x$$ < $$20$$
x^2-x-20 < 0$$x^{2}-x-20$$ < $$0$$
Paaiškinimas: (x-5)* (x+4) < 0$$(x-5)\cdot (x+4)$$ < $$0$$
x-5 = 0$$x-5$$ = $$0$$
x = 0+5$$x$$ = $$0+5$$
x+4 = 0$$x+4$$ = $$0$$
x = 0-4$$x$$ = $$0-4$$
Lygties sprendiniai x = - 4 ir x = 5.
Intervalai (-∞; - 4), (- 4; 5) ir (5; +∞).
$$(x-5)\cdot (x+4) < 0$$, kai x priklauso intervalui (- 4; 5).
Pradinės nelygybės apibrėžimo sritis
$$x-1 > 0$$
$$x > 1$$
Apjungus intervalą su apibrėžimo sritimi, gauname x priklauso (1; 5)
Atsakymas: (1; 5)