Відповідь:
19x-11y+12=0
Покрокове пояснення:Simplifying
19x + -11y + 12 = 0
Reorder the terms:
12 + 19x + -11y = 0
Solving
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-12' to each side of the equation.
12 + 19x + -12 + -11y = 0 + -12
12 + -12 + 19x + -11y = 0 + -12
Combine like terms: 12 + -12 = 0
0 + 19x + -11y = 0 + -12
19x + -11y = 0 + -12
Combine like terms: 0 + -12 = -12
19x + -11y = -12
Add '11y' to each side of the equation.
19x + -11y + 11y = -12 + 11y
Combine like terms: -11y + 11y = 0
19x + 0 = -12 + 11y
19x = -12 + 11y
Divide each side by '19'.
x = -0.6315789474 + 0.5789473684y
Simplifying
(0;2]U[4;6)
Пошаговое объяснение:
ОДЗ:
{x > 0;
{6–x > 0 ⇒ x < 6
{(x4–12x3+36x2) > 0⇒ (x·(6–x))2 > 0 ⇒ x≠0; x≠6
ОДЗ: х∈(0;6)
при х∈(0;6):
log2(x4–12x3+36x2)=log2x2·(6–x)2=
log2(x·(6–x))2=2log2x·(6–x)=2log2x+2log2(6–x)
Неравенство принимает вид:
(2–log2x)·(log2(6–x)–2) ≥ 0
Применяем обобщенный метод интервалов
log2x=2 или log2(6–x)=2
x=4 или 6–х=4;х=2
При х=1
(2–log21)·(log2(6–1)–2)=2·(log25–log24) > 0
При х=3
(2–log23)·(log2(6–3)–2)=–(2–log23)2 < 0
При х=5
(2–log25)·(log2(6–5)–2)=(log24–log25)·(0–2) > 0
(0)__+__ [2]__–__[4]__+__ (6)
Відповідь:
19x-11y+12=0
Покрокове пояснення:Simplifying
19x + -11y + 12 = 0
Reorder the terms:
12 + 19x + -11y = 0
Solving
12 + 19x + -11y = 0
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-12' to each side of the equation.
12 + 19x + -12 + -11y = 0 + -12
Reorder the terms:
12 + -12 + 19x + -11y = 0 + -12
Combine like terms: 12 + -12 = 0
0 + 19x + -11y = 0 + -12
19x + -11y = 0 + -12
Combine like terms: 0 + -12 = -12
19x + -11y = -12
Add '11y' to each side of the equation.
19x + -11y + 11y = -12 + 11y
Combine like terms: -11y + 11y = 0
19x + 0 = -12 + 11y
19x = -12 + 11y
Divide each side by '19'.
x = -0.6315789474 + 0.5789473684y
Simplifying
x = -0.6315789474 + 0.5789473684y
(0;2]U[4;6)
Пошаговое объяснение:
ОДЗ:
{x > 0;
{6–x > 0 ⇒ x < 6
{(x4–12x3+36x2) > 0⇒ (x·(6–x))2 > 0 ⇒ x≠0; x≠6
ОДЗ: х∈(0;6)
при х∈(0;6):
log2(x4–12x3+36x2)=log2x2·(6–x)2=
log2(x·(6–x))2=2log2x·(6–x)=2log2x+2log2(6–x)
Неравенство принимает вид:
(2–log2x)·(log2(6–x)–2) ≥ 0
Применяем обобщенный метод интервалов
log2x=2 или log2(6–x)=2
x=4 или 6–х=4;х=2
При х=1
(2–log21)·(log2(6–1)–2)=2·(log25–log24) > 0
При х=3
(2–log23)·(log2(6–3)–2)=–(2–log23)2 < 0
При х=5
(2–log25)·(log2(6–5)–2)=(log24–log25)·(0–2) > 0
(0)__+__ [2]__–__[4]__+__ (6)