3
1
x
2
+x−
3
10
<0 ⇒ x
2
+3x−10<0 ,
D=9+40=49=7
2
, x
1
=−5 , x
2
=2
(x+5)(x−2)<0
znaki: +++(−5)−−−(2)+++
x∈(−5 ;2 )
2) x
2
+10x+25>0 , (x+5)
2
>0 → x+5
=0 , x
=−5
x∈(−∞;−5 )∪(−5 ;+∞)
3) 3x
2
−24x+48<0 , x
2
−8x+16<0 , (x−4)
2
<0 ,
x∈∅
\begin{gathered}4)\ \ x^2+\dfrac{2}{3}\, x+\dfrac{4}{3} > 0\ \ \ ,\ \ \ 3x^2+2x+4 > 0\ \ ,D/4=1-12=-11 < 0\ \ \Rightarrow \ \ \ x\in \varnothing 5)\ \ -4x^2+5x-2 > 0\ \ \ ,\ \ \ 4x^2-5x+2 < 0\ \ ,\ \ D=25-32=-7 < 0\ ,x\in \varnothing\end{gathered}
4) x
2
+
3
2
x+
3
4
>0 , 3x
2
+2x+4>0 ,
D/4=1−12=−11<0 ⇒ x∈∅
5) −4x
2
+5x−2>0 , 4x
2
−5x+2<0 , D=25−32=−7<0 ,
x∈∅
y` = 4x^3 +6x
y` = 3x^2-6x+1
y`= 6x+2
y`= 4x+ 1/ cos^2 x
y` = 5x^4-10x + cosx
y`= e^x + 1/x
y`= 1- 1/x
y`= -sinx +cos x
y`= 1/ (2*корень из х) - 1/ (х^2)
y`= 1/ (x ln 7) + 3
y`= 1/ (x ln 3) + 1/ (x ln 5)
y`= 5+2=7
y`= [(2x+5)(2-8x)+8(x^2+5x)] / (2-8x)^2 = (-8x^2+4x+10) / (2-8x)^2
y`= 6x
y`=9x^2-6
y`= cosx(1+cosx) - sinx(1+sinx)= cosx+cos^2 x-sinx-sin^2 x= cosx - sinx+ cos2x
y`= 1/( cos^2 x) - 2cosx
y`= 12x^2
y`= 12x^2-8
y`= 1/x * (x^2-1)+2x*lnx=(x^2-1) / x + 2x*lnx
y`= 4^x * ln4 * log4x + 4^x / (x*ln4)
