(x+5)²>x(x+10)
x²+10x+25>x²+10x
25>0
всегда
<span>1)sin(x-pi/4)=0</span>
t=<span>x-pi/4</span>
<span>sint=0</span>
<span>1.t=0+2pi*k</span>
x-pi/4=0
x=pi/4
2.t=pi+2pi*k
pi=x-pi/4
x=5pi/4+2pi*k
3)<span>sinx(3x+pi/3)=1</span>
<span>t=<span>3x+pi/3</span></span>
<span><span>sint=1</span></span>
<span><span>t=pi/2</span></span>
<span><span>3x+pi/3=pi/2</span></span>
<span><span>x=pi/18</span></span>
<span><span>5)<span>tg(x-pi/6)=-√3</span></span></span>
<span><span><span>t=x-pi/6</span></span></span>
<span><span><span>tgt=-√3</span></span></span>
<span><span><span>t=-pi/3+2pi*k . k=Z</span></span></span>
<span><span><span>t=2pi/3+2pi*k . k=Z</span></span></span>
1) log(√7)(1/7)+3^log₃7=log(7^(1/2)7⁻¹+7=-2log₇7+7=-2+7=5.
2) √(25^(1/log⁶5)+49^(1/log₈7))=√(5^2log₅6+7^2log₇8)=
=√(5^log₅6²+7^log₇8²)=√(6²+8²)=√100=10.
3) 10^(1-lg5)=10*10^(-lg5)=10/10^lg5=10/5=2.
4) log(c)(16c²)=log(c)4⁴+log(c)c²=4log(c)2+2=4*(-3)+2=-10.
(5^2х25^2)/5^8 = (5^2х5^3)/5^8 = 5^8/5^8 = 1
Решение
у = √(х² - 64)
х² - 64 ≥ 0
x² = 64
x₁ = - 8
x₂ = 8
область определения функции: x ∈ ( - ∞; - 8]∪[-3;3]8; + ∞)
у=√х+3+√3-х
x + 3 ≥ 0
3 - x ≥ 0
x ≥ - 3
x ≤ 3
<span>область определения функции:</span>
x ∈ [- 3;3]