1) log₂ 5=<u> 1 </u>
log₅ 2
2) <u> 1 </u> + 16log₅ 2 - 8 = <u>1 + 16log₅² 2 - 8log₅ 2</u>
log₅ 2 log₅ 2
3) <u> 1+16log₅² 2 - 8log₅ 2 </u> * log₅ 2 = 1+16log₅² 2 - 8log₅ 2 =
log₅ 2
= 16log₅² 2 - 8log₅ 2 + 1= (4log₅ 2 - 1)²
4) √(4log₅ 2 -1)² = |4log₅ 2 - 1| = 4log₅ 2 - 1 = log₅ 2⁴ - 1
5) 4log₅ 12.5 = 4log₅ (25/2) = 4(log₅ 25 - log₅ 2) =
= 4(2 - log₅ 2) = 8 - 4log₅ 2 = 8 - log₅ 2⁴
6) log₅ 2⁴ - 1 + 8 - log₅ 2⁴ = 7
Ответ: 7
Держи!Вроде так..............
D=b^2 - 4ac = ( -3)^2 - 4*4*(-37)=9+592=601
√D=√601
x1=3+√ 601 / 8
x2= 3 - √ 601 / 8
4х²+20х=0 12-48х₂=0 3х²-15=0 2х²+32=0 5х²-7х+2=0
х(4х+20)=0 -48х²=-12 3х²=15 2х²=-32 Д=49-4*5*2=9
4х+20=0 х²=-12/-48 х²=5 х²=-16 х₁=7-3/10=2/5
4х=-20 х²=0.25 х=+-√5 х=+-4 x₂=7+3/10=1
х=-5 х=+-0.5