Х^2+7х-10х-70=0
х^2-3х-70=0
D=(-3)^2-4*1*(-70)=9+280=289>0=>√289=17
х1=3-17/2=-14/2=-7
х2=3+17/2=20/2=10
Надеюсь правильно(´・・`)
<span>2x>|x|+1
2x-|x|>1
2x-x>1, x</span>≥0
2x-(-x)>1, x<0
x>1, x≥0
x>1/3, x<0
x∈(1,+∞)
x∈∅
Ответ: <span>x∈(1,+∞)</span>
-2^3/(-2+1/3)+1/3=-8/(-2)+(-8/1/3)=4-8*3/1=4-24=-20
1) 8^(1/3) = 2.
2* log2 (6) = log2 (9*4) = log2 (9) + 2
2) укажите основание у логарифма
3) (1/2)^(1/2) = 1/кореньиз(2)
log2 (4) /кореньиз (2) + 1 = 1+корень (2)
4) = log3 (3^(-1/4)* 3^2) = log3 (3^(7/4)) = 7/4
5) = log9 (27) = log3 (3^(3/2) = 3/2
Вроде бы так.
А)=2(а²-6а+9)=2(а-3)²
б)=-5(х²-2х+1)=-5(х-1)²
в)=0,5u²+2uv+(2uv+8v²)=
u(0,5u+2v)+2v(u+4v)=u/2(u+4v)+2v(u+4v)
=(u+4v)(u/2+2v)=1/2(u+4v)+u+4v)
г)=-1/10(9х²+6ху+у²)=-1/10(3х+у)²
д)=t(t²-8t+16)=t(t-4)²
e)=u(4u²+4u+1)=u(2u+1)²