Решение
log₂ 6 - log₂ (3/16) = log₂ [6 / (3/16)] = log₂ 32 = log₂ 2⁵ = 5log₂ 2 = 5
Σ₂₀ а₂₀=7 а₂-?
Σ₂₀=(а₁+а₂₀)*n/2=(a₁+7)*20/2=10a₁+70=240
10a₁=170
a₁=17
a₂₀=a₁+19d=17+19d=7
19d=-10
d=-10/19
a₂=a₁+d=17+(-10/19)=16_9/19.
a₂=16_9/19.
5х-у-2=0
х²-2ху+у²<span>=4
-y=2-5x
</span>х²-2ху+у²=4
y=5x-2
x²-2x*(5x-2)+(5x-2)²=4
x²-10x²+4x+25x²-20x+4-4=0
16x²-16x=0
16x(x-1)=0
16x=0 или x-1=0
x1=0 x2=1
_____ ________
y=5x-2;
y1=5x1-2=5*0-2=0-2=-2
y2=5x2-2=5*1-2=5-2=3
Ответ: (0 ; -2) ; (1 ; 3)