lJNln
Išspręskite nelygybę: $$5^{(2*x+1)}-5^{(x+2)}$$ <= $$5^{(x+1)}-25$$
Sprendimas.
5^(2*x+1)- 5^(x+2) ≤
5^(x+1)-25
5^(x+1)-25
|
5^(2*x+1)- 5^(x+2) ≤ 5^(x+1)-25$$5^{(2*x+1)}-5^{(x+2)}$$ ≤ $$5^{(x+1)}-25$$
5^(2*x)* 5- 5^(x+2) ≤ 5^(x+1)-25$$5^{(2*x)}\cdot 5-5^{(x+2)}$$ ≤ $$5^{(x+1)}-25$$
5^(2*x)* 5- 5^x* 25 ≤ 5^(x+1)-25$$5^{(2*x)}\cdot 5-5^{x}\cdot 25$$ ≤ $$5^{(x+1)}-25$$
5^(2*x)* 5- 5^x* 25 ≤ 5^x* 5-25$$5^{(2*x)}\cdot 5-5^{x}\cdot 25$$ ≤ $$5^{x}\cdot 5-25$$
5^(2*x)* 5- 5^x* 25- 5^x* 5 ≤ -25$$5^{(2*x)}\cdot 5-5^{x}\cdot 25-5^{x}\cdot 5$$ ≤ $$-25$$
5^(2*x)* 5- 5^x* 25- 5^x* 5+25 ≤ 0$$5^{(2*x)}\cdot 5-5^{x}\cdot 25-5^{x}\cdot 5+25$$ ≤ $$0$$
$${\normalsize -5^{x}\cdot 25-5^{x}\cdot 5}$$ = $${\normalsize -30\cdot 5^{x}}$$
5^(2*x)* 5- 30* 5^x+25 ≤ 0$$5^{(2*x)}\cdot 5-30\cdot 5^{x}+25$$ ≤ $$0$$
Keitimas $${\normalsize 5^{x} = a}$$
a^2* 5- 30* 5^x+25 ≤ 0$$a^{2}\cdot 5-30\cdot 5^{x}+25$$ ≤ $$0$$
Keitimas $${\normalsize 5^{x} = a}$$
a^2* 5- 30* a+25 ≤ 0$$a^{2}\cdot 5-30\cdot a+25$$ ≤ $$0$$
$${\normalsize a^{2}\cdot 5-30\cdot a+25}$$ = $${\normalsize 5\cdot (a^{2}-6\cdot a+5)}$$ Paaiškinimas:
Paaiškinimas: 5 iškeltas prieš skliaustus
5* ( a^2- 6* a+5) ≤ 0$$5\cdot (a^{2}-6\cdot a+5)$$ ≤ $$0$$
$${\normalsize a^{2}-6\cdot a+5}$$ = $${\normalsize (a-5)\cdot (a-1)}$$ Paaiškinimas:
Paaiškinimas: Kvadratinis trinaris $${\normalsize a\cdot x^{2}+b\cdot x+c}$$, kur
a = 1, b = -6, c = 5.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 36-20}$$ = 16.
x1 = $${\normalsize \frac{6+\sqrt {16}}{2\cdot 1} = \frac{6+4}{2} = \frac{3+2}{1} = \frac{5}{1}}$$ = 5
x2 = $${\normalsize \frac{6-\sqrt {16}}{2\cdot 1} = \frac{6-4}{2} = \frac{3-2}{1} = \frac{1}{1}}$$ = 1
a = 1, b = -6, c = 5.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 36-20}$$ = 16.
x1 = $${\normalsize \frac{6+\sqrt {16}}{2\cdot 1} = \frac{6+4}{2} = \frac{3+2}{1} = \frac{5}{1}}$$ = 5
x2 = $${\normalsize \frac{6-\sqrt {16}}{2\cdot 1} = \frac{6-4}{2} = \frac{3-2}{1} = \frac{1}{1}}$$ = 1
5* ( (a-5)* (a-1)) ≤ 0$$5\cdot ((a-5)\cdot (a-1))$$ ≤ $$0$$
$${\normalsize 5\cdot ((a-5)\cdot (a-1))}$$ = $${\normalsize 5\cdot (a-5)\cdot (a-1)}$$
5* (a-5)* (a-1) ≤ 0$$5\cdot (a-5)\cdot (a-1)$$ ≤ $$0$$
Gavosi kvadratinė nelygybė $$a^{2}-6\cdot a+5$$ <= 0, parabolės šakos nukreiptos į viršų, a ašį kerta taškuose a = 1, ir a = 5,
todėl a >= 1 ir a <= 5, t.y. a priklauso intervalui [1; 5].
Buvo atliktas keitimas $$5^{x} = a$$, gauname dvi nelygybes:
$$5^{x}$$ >= $$1$$ ir $$5^{x}$$ <= $$5$$
Iš pirmos nelygybės gauname
$$5^{x}$$ >= $$1$$
$$5^{x}$$ >= $$5^{0}$$
x >= 0
Iš antros nelygybės gauname
$$5^{x}$$ <= $$5$$
$$5^{x}$$ <= $$5^{1}$$
x <= 1
Atsakymas. x >= 0 ir x <= 1 arba x priklauso intervalui [0; 1]