1BzCs
Laikydami, kad x > 0, raskite x iš lygybės:
$$5^{2}\cdot 5^{4}\cdot 5^{6}$$ * ... * $$5^{(2*x)} = 0.04^{(-28)}$$
Sprendimas.
Pertvarkome dešinę pusę, kad gautume 5 laipsnį:
5^(2+4+6 .+ ... + 2x) = $$(\frac{1}{25})^{(-28)}$$
5^(2+4+6 + ... + 2x) = $$((\frac{1}{5})^{2})^{(-28)}$$
5^(2+4+6 + ... + 2x) = $$((5)^{(-2)})^{(-28)}$$
5^(2+4+6 +... + 2x) = $$5^{56}$$
2+4+6 + ... + 2x = 56
Kairėje pusėje aritmetinė progresija, kurioje yra x narių, pirmasis 2, x-asis 2x.
Pagal aritmetinės progresijos sumos formulę $$S_{n} = \frac{(a_{1}+a_{n})\cdot n}{2}$$, gauname
$$\frac{(2+2\cdot x)\cdot x}{2} = 56$$
(1 + x) x = 56
(1+x)* x =
56
56
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(1+x)* x = 56$$(1+x)\cdot x$$ = $$56$$
$${\normalsize (1+x)\cdot x}$$ = $${\normalsize 1\cdot x+x\cdot x}$$
1* x+ x* x = 56$$1\cdot x+x\cdot x$$ = $$56$$
$${\normalsize x\cdot x}$$ = $${\normalsize x^{2}}$$
1* x+ x^2 = 56$$1\cdot x+x^{2}$$ = $$56$$
1* x+ x^2-56 = 0$$1\cdot x+x^{2}-56$$ = $$0$$
x^2+ 1* x-56 = 0$$x^{2}+1\cdot x-56$$ = $$0$$
$${\normalsize 1\cdot x}$$ = $${\normalsize x}$$
x^2+x-56 = 0$$x^{2}+x-56$$ = $$0$$
$${\normalsize x^{2}+x-56}$$ = $${\normalsize (x-7)\cdot (x+8)}$$ Paaiškinimas:
Paaiškinimas: Kvadratinis trinaris $${\normalsize a\cdot x^{2}+b\cdot x+c}$$, kur
a = 1, b = 1, c = -56.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 1-(-224)}$$ = 225.
x1 = $${\normalsize \frac{-1+\sqrt {225}}{2\cdot 1} = \frac{-1+15}{2} = \frac{14}{2}}$$ = 7
x2 = $${\normalsize \frac{-1-\sqrt {225}}{2\cdot 1} = \frac{-1-15}{2} = \frac{-16}{2}}$$ = -8
a = 1, b = 1, c = -56.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 1-(-224)}$$ = 225.
x1 = $${\normalsize \frac{-1+\sqrt {225}}{2\cdot 1} = \frac{-1+15}{2} = \frac{14}{2}}$$ = 7
x2 = $${\normalsize \frac{-1-\sqrt {225}}{2\cdot 1} = \frac{-1-15}{2} = \frac{-16}{2}}$$ = -8
(x-7)* (x+8) = 0$$(x-7)\cdot (x+8)$$ = $$0$$
Gavome dvi šaknis, teigiama tik viena x = 7
Atsakymas: x = 7