Очень прошу , решите неравенства с 1 по 5 даю высокий бал

Question

Дан 1 ответ

flsh_zn · Answer 1 · 2018-04-18T17:25:29+0000

1.
$\sqrt{x} \ \textless \ \sqrt[4]{6+x} \\ \\ \left \{x^2 \ \textless \ 6+x} \atop {x \geq 0}} \right. \\ \\ \left \{x^2-x-6 \ \textless \ 0} \atop {x \geq 0}} \right.\\ \\ \left \{ -2\ \textless \ x\ \textless \ 3} \atop {x \geq 0}} \right.$
x ∈ [0; 3)
2.
$\sqrt{3x+1}\ \textgreater \ \sqrt[3]{7x+1}$
(3x + 1)³ > (7x + 1)²
27x³ + 27x² + 9x + 1 > 49x² + 14x + 1
27x³ - 22x² - 5x > 0
x(27x² - 22x - 5) > 0
x(x - 1)(27x + 5) > 0
x ∈ (-5/27; 0) ∪ (1; +∞)
3.
$\frac{x^2}{1-cos \frac{ \pi x}{2} } \ \textless \ \frac{8x-12}{1-cos \frac{ \pi x}{2} } \\ \\ \frac{x^2-8x+12}{1-cos \frac{ \pi x}{2} } \ \textless \ 0 \\ \\\frac{(x-2)(x-6)}{1-cos \frac{ \pi x}{2} } \ \textless \ 0 \\ \\ \left \{ {(x-2)(x-6)\ \textless \ 0} \atop {cos \frac{ \pi x}{2} \neq 1}} \right. \\ \\ \left \{ {{xE(2; 6)} \atop {\frac{ \pi x}{2} \neq 2 \pi n,nEZ}} \right. \\ \\\left \{ {{xE(2; 6)} \atop {x\neq 4n,nEZ}} \right.$
x ∈(2; 4) ∪ (4; 6)
4.
$\frac{2cosx}{ \sqrt{2-x-x^2} } \ \textgreater \ \frac{1}{ \sqrt{2-x-x^2} } \\ \\ \frac{2cosx-1}{ \sqrt{2-x-x^2} } \ \textgreater \ 0 \\ \\ \left \{ {{2cosx-1\ \textgreater \ 0} \atop {2-x-x^2\ \textgreater \ 0}} \right.\\ \\ \left \{ {{2cosx\ \textgreater \ 1} \atop {x^2+x-2\ \textless \ 0}} \right.\\ \\ \left \{ {{cosx\ \textgreater \ \frac{1}{2} } \atop {(x+2)(x-1)\ \textless \ 0}} \right.\\ \\ \left \{ {{xE(- \frac{ \pi }{3}+2 \pi n; \frac{ \pi }{3}+2 \pi n), nEZ} \atop {xE(-2; 1)} \right.$
x ∈(-π/3; 1)
5.
$log_4^2x\ \textless \ \frac{2}{log_x4} \\ \\log_4^2x\ \textless \ 2*log_4x}$
log₄x·(log₄x - 2) < 0
Замена: log₄x = t
t·(t - 2) < 0
t ∈(0; 2)
x ∈(1; 16)
6.
√2·tg x·ctg x < 2· cos x
$\left \{ {{ \frac{ \sqrt{2}}{2}\ \textless \ cosx} \atop {x \neq \frac{ \pi n}{2},nEZ }} \right. \\ \\ \left \{ {{xE(- \frac{ \pi }{4}+2 \pi k; \frac{ \pi }{4}+2 \pi k), kEZ} \atop {x \neq \frac{ \pi n}{2},nEZ }} \right.$
x ∈(-π/4 + 2πn; 2πn) ∪ (2πn; π/4 + 2πn), n ∈ Z