ОДЗ: x + 1 > 0 ⇒ x > - 1
x + 2 > 0 ⇒ x > - 2
Значит окончательно: x ∈ (- 1 ; + ∞)
![log _{2} (x+1)+log _{2} (x+2)=3-log _{2}4\\\\log _{2} (x+1)(x+2)=log_{2}8 -log _{2}4\\\\log _{2}(x+1)(x+2)=log _{2} \frac{8}{4} \\\\log _{2}(x+1)(x+2)=log _{2} 2\\\\(x+1)(x+2)=2\\\\ x^{2} +2x+x+2-2=0\\\\ x^{2} +3x=0\\\\ x(x+3)=0\\\\ x_{1} =0\\\\x+3=0\\\\ x_{2}=-3](https://tex.z-dn.net/?f=log+_%7B2%7D+%28x%2B1%29%2Blog+_%7B2%7D+%28x%2B2%29%3D3-log+_%7B2%7D4%5C%5C%5C%5Clog+_%7B2%7D+%28x%2B1%29%28x%2B2%29%3Dlog_%7B2%7D8+-log+_%7B2%7D4%5C%5C%5C%5Clog+_%7B2%7D%28x%2B1%29%28x%2B2%29%3Dlog+_%7B2%7D+%5Cfrac%7B8%7D%7B4%7D+%5C%5C%5C%5Clog+_%7B2%7D%28x%2B1%29%28x%2B2%29%3Dlog+_%7B2%7D+2%5C%5C%5C%5C%28x%2B1%29%28x%2B2%29%3D2%5C%5C%5C%5C+x%5E%7B2%7D+%2B2x%2Bx%2B2-2%3D0%5C%5C%5C%5C+x%5E%7B2%7D+%2B3x%3D0%5C%5C%5C%5C+x%28x%2B3%29%3D0%5C%5C%5C%5C++x_%7B1%7D+%3D0%5C%5C%5C%5Cx%2B3%3D0%5C%5C%5C%5C+x_%7B2%7D%3D-3+++++++)
x₂ = - 3 - не подходит
Ответ : 0
25x²-10x+1=0
D=-10²-4*25*1=100-100=0
x=10±0/2*25=10/50=1/5=0,2
25*(0,2)²-10*0,2+1=0
Ответ:x=0,2
F'(x)=6x^2+5x-1 формула (x^n)'=nx^(n-1)
f'(x)>0
6x^2+5x-1>0 6x^2+5x-1=0 a+c=b⇒x1=-1 x2=-c/a=1/6
(x+1)(x-1/6)>0
+ -1 - 1/6 +
ответ (-00,-1)∪(1/6,+00)