x⁴+2018x²-2019=0
Пусть x²=t≥0 ⇒
t²+2018t-2019=0
D=2018²-4*(-2019)=4072324+8076=4080400
√D=√4080400=2020 ⇒
t₁=1 t₂=-2019 ∉ ⇒
x²=1
x₁=1 x₂=-1.
Ответ: x₁=1, x₂=-1.
10,2(16-2)+19,8(16-2)=10,2*14+19,8*14=14(10,2+19,8)=14*30=420
А) (m¼ - n½)² - (m¼ + n½)²=m^1/2-2(mn)^1/4+n^1/2+m^1/2+2(mn)^1/4+n^1/2=
=2√m+2√n<span>
б) (m⅓ + 3n½)² + (m⅓ - 3n½)²=
=m^2/3+6m^1/3n^1/2+9n+</span>m^2/3-6m^1/3n^1/2+9n=2∛m²+18n<span>
в) (m½ - 2n¼) (m½ + 2n¼)=m-4</span>√n<span>
г) (m½ - 3n) (m + 3m½n + 9n²)=m^3/2-27n</span>³=√m³-27n³
С^2+8с+16=0 →два корня
D=64-64=0 - один корень
с=-4