Нужно заучить таблицу квадратов, хотя бы до 20^2 = 400, и извлекать корни.
1) √(16*25) = √16*√25 = 4*5 =. 20
2) √216 = √(36*6) = 6√6
3) 2√14 = √(2^2*14) = √(4*14) = √56
4) 6+ 4√2 = 4+ 4√2+ 2. = 2^2+ 2*2*√2+ (√2)^2 = (2+ √2)^2
5) 26 -15√3 = 8 +18 - 12√3 -3√3 = 2^3 - 3*2^2*√3 + 3*2*(√3)^2 - (√3)^3 = (2-√3)^3
6) (√2-1)*√(3-2√2) + 2√2 = (√2-1)*√(2-2√2+1) + 2√2 = (√2-1)*√(√2-1)^2 + 2√2 =
= (√2-1)(√2-1) + 2√2 = 2-2√2+1+2√2 = 3
7) 12/(3√2) = 6/√2 = 6*√2/(√2)^2 = 6√2/2 = 3√2
8) 4/(3-√15) + 4/(3+√15) = 4(3+√15)/(9-15) + 4(3-√15)/(9-15)=
= (12+4√15+12-4√15)/(-6) = 24/(-6) = -4
Решить неравенство
<span>(m^2-3m-2)(m^2-3m-3) ≤ 2</span>
Решение
<span>Пусть z = m² - 3m, тогда<span>
(m</span></span>²<span><span>-3m-2)(m</span></span>²<span><span>-3m-3)
= (z - 2) * (z - 3) = z² - 5z + 6
</span>z² - 5z + 6 ≤ 2<span>
z² - 5z + 4 ≤ 0
z</span></span>₁<span><span> = 1
z</span></span>₂<span><span> = 4
1) m</span></span>²<span><span> – 3m = 1
m</span></span>²<span><span> – 3m – 1 = 0
D = 9 + 4*1*1 = 13
m</span></span>₁<span><span> = (3 - √13)/2
m</span></span>₂<span><span> = (3 + √13)/2
<span>2) m</span></span></span>²<span><span><span> – 3m = 4</span>
m</span></span>²<span><span> – 3m – 4 = 0
m</span></span>₃<span><span> = - 1
<span>m</span></span></span>₄<span><span><span> = 4
+ - + - +
---------------------------------------------------------------------------------------->
</span></span></span>(3 - √13)/2 - 1 (3 + √13)/2 4 x
m ∈ [ (3 - √13)/2 ; - 1] [ (3 + √13)/2 ; 4]
Это Элементарно это будет 3
60*3=180км - проехали на машине
180+12=192 км - весь путь
Третья сторона не должна быть больше 13 и меньше 3