ОДЗ 121-х²≥0
(11-х)(11+х)≥0
х=11 х=-11
х∈[-11;11]
(х+1)/(x²-17x+16)≥0
x+1=0⇒x=-1
x²-17x+16=0
x1+x2=17 U x1*x2=16
x1=1 U x2=16
_ + _ +
-----------[-1]----------(1)-------------(16)------------------
-1≤x<1 U x>16+ОДЗ⇒x∈[-1;1)
!!!!!!!!!!!!!!!!!!!!!!!!!!
5•40+8•8-5•40²/40
5•40+64-5•40
200+64-200=64
㏒ₐа⁶*b¹⁰=㏒ₐа⁶+㏒ₐb¹⁰=6㏒ₐа+10㏒ₐb=6+10*8=86
㏒ₐа⁶/b⁴=㏒ₐа⁶ -㏒ₐb⁴=6㏒ₐа -4㏒ₐb=6-4*(-2)=14