Проведите 2 высоты нижнее основание разделилось на отрезки 3см,8см, 3см
Из Δавк ав²=ак²+вк²=3²+2²=13
ав=√13
x2 + 12x + 33 = 0
D = b2 - 4ac
D = 144 - 132 = 12 = (2√3)^2
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Известная формула:
То есть значения "х" располагаются левее (-а) и правее (а) .
Вместо "х" может быть записано любое выражение, а вместо "а" - любое число. В формуле надо только заменить "х" и "а" на те выражения или числа, которые заданы в условии.
![201)\; \; \Big |\frac{3x+1}{x-5}\Big |\geq 1\; \; \to \; \; \left [ {{\frac{3x+1}{x-5}\geq 1} \atop {\frac{3x+1}{x-5}\leq -1}} \right. \\\\a)\; \; \frac{3x+1}{x-5}-1\geq 0\; ,\; \; \frac{3x+1-(x-5)}{x-5}\geq 0\; ,\; \; \frac{2x+6}{x-5}\geq 0\; ,\; \; \frac{2(x+3)}{x-5}\geq 0\; ,\\\\+++[-3]---(5)+++\\\\x\in (-\infty ,-3\; ]\cup (5,+\infty )\\\\b)\; \; \frac{3x+1}{x-5}+1\leq 0\; ,\; \; \frac{3x+1+x-5}{x-5} \leq 0\; ,\; \; \frac{4x-4}{x-5}\leq 0\; ,\; \frac{4(x-1)}{x-5}\leq 0\; ,\\\\+++[\; 1\; ]---(5)+++](https://tex.z-dn.net/?f=201%29%5C%3B+%5C%3B+%5CBig+%7C%5Cfrac%7B3x%2B1%7D%7Bx-5%7D%5CBig+%7C%5Cgeq+1%5C%3B+%5C%3B+%5Cto+%5C%3B+%5C%3B+%5Cleft+%5B+%7B%7B%5Cfrac%7B3x%2B1%7D%7Bx-5%7D%5Cgeq+1%7D+%5Catop+%7B%5Cfrac%7B3x%2B1%7D%7Bx-5%7D%5Cleq+-1%7D%7D+%5Cright.+%5C%5C%5C%5Ca%29%5C%3B+%5C%3B+%5Cfrac%7B3x%2B1%7D%7Bx-5%7D-1%5Cgeq+0%5C%3B+%2C%5C%3B+%5C%3B+%5Cfrac%7B3x%2B1-%28x-5%29%7D%7Bx-5%7D%5Cgeq+0%5C%3B+%2C%5C%3B+%5C%3B+%5Cfrac%7B2x%2B6%7D%7Bx-5%7D%5Cgeq+0%5C%3B+%2C%5C%3B+%5C%3B+%5Cfrac%7B2%28x%2B3%29%7D%7Bx-5%7D%5Cgeq+0%5C%3B+%2C%5C%5C%5C%5C%2B%2B%2B%5B-3%5D---%285%29%2B%2B%2B%5C%5C%5C%5Cx%5Cin+%28-%5Cinfty+%2C-3%5C%3B+%5D%5Ccup+%285%2C%2B%5Cinfty+%29%5C%5C%5C%5Cb%29%5C%3B+%5C%3B+%5Cfrac%7B3x%2B1%7D%7Bx-5%7D%2B1%5Cleq+0%5C%3B+%2C%5C%3B+%5C%3B+%5Cfrac%7B3x%2B1%2Bx-5%7D%7Bx-5%7D+%5Cleq+0%5C%3B+%2C%5C%3B+%5C%3B+%5Cfrac%7B4x-4%7D%7Bx-5%7D%5Cleq+0%5C%3B+%2C%5C%3B+%5Cfrac%7B4%28x-1%29%7D%7Bx-5%7D%5Cleq+0%5C%3B+%2C%5C%5C%5C%5C%2B%2B%2B%5B%5C%3B+1%5C%3B+%5D---%285%29%2B%2B%2B)
![x\in [\; 1,5)\\\\c)\; \; \left [ {{x\in (-\infty ,-3\, ]\cup (5,+\infty )} \atop {x\in [\, 1,5)}} \right. \; \; \Rightarrow \; \; \underline {x\in (-\infty ,-3\, ]\cup [\, 1,5)\cup (5,+\infty )}\\\\\\203)\; \; |3x-2|>2x+1\; \; \Rightarrow \; \; \left \{ {{3x-2>2x+1} \atop {3x-2<-(2x+1)}} \right. \\\\a)\; \; 3x-2>2x+1\; ,\; \; x>3\\\\b)\; \; 3x-2<-(2x+1)\; ,\; \; 3x-2<-2x-1\; ,\; \; 5x<1\; ,\; x<\frac{1}{5}\\\\c)\; \; \left [ {{x>3} \atop {x<\frac{1}{5}}} \right. \; \; \Rightarrow \; \; \underline {x\in (-\infty ,\frac{1}{5})\cup (3,+\infty )}](https://tex.z-dn.net/?f=x%5Cin+%5B%5C%3B+1%2C5%29%5C%5C%5C%5Cc%29%5C%3B+%5C%3B+%5Cleft+%5B+%7B%7Bx%5Cin+%28-%5Cinfty+%2C-3%5C%2C+%5D%5Ccup+%285%2C%2B%5Cinfty+%29%7D+%5Catop+%7Bx%5Cin+%5B%5C%2C+1%2C5%29%7D%7D+%5Cright.+%5C%3B+%5C%3B+%5CRightarrow+%5C%3B+%5C%3B+%5Cunderline+%7Bx%5Cin+%28-%5Cinfty+%2C-3%5C%2C+%5D%5Ccup+%5B%5C%2C+1%2C5%29%5Ccup+%285%2C%2B%5Cinfty+%29%7D%5C%5C%5C%5C%5C%5C203%29%5C%3B+%5C%3B+%7C3x-2%7C%3E2x%2B1%5C%3B+%5C%3B+%5CRightarrow+%5C%3B+%5C%3B+%5Cleft+%5C%7B+%7B%7B3x-2%3E2x%2B1%7D+%5Catop+%7B3x-2%3C-%282x%2B1%29%7D%7D+%5Cright.+%5C%5C%5C%5Ca%29%5C%3B+%5C%3B+3x-2%3E2x%2B1%5C%3B+%2C%5C%3B+%5C%3B+x%3E3%5C%5C%5C%5Cb%29%5C%3B+%5C%3B+3x-2%3C-%282x%2B1%29%5C%3B+%2C%5C%3B+%5C%3B+3x-2%3C-2x-1%5C%3B+%2C%5C%3B+%5C%3B+5x%3C1%5C%3B+%2C%5C%3B+x%3C%5Cfrac%7B1%7D%7B5%7D%5C%5C%5C%5Cc%29%5C%3B+%5C%3B+%5Cleft+%5B+%7B%7Bx%3E3%7D+%5Catop+%7Bx%3C%5Cfrac%7B1%7D%7B5%7D%7D%7D+%5Cright.+%5C%3B+%5C%3B+%5CRightarrow+%5C%3B+%5C%3B+%5Cunderline+%7Bx%5Cin+%28-%5Cinfty+%2C%5Cfrac%7B1%7D%7B5%7D%29%5Ccup+%283%2C%2B%5Cinfty+%29%7D)
"х" принимает значения из промежутка между (-а) и (а) .
A) 2sinαcosα , α =π/2.
2sinαcosα =sin2α =sin(2*π/2) =sinπ = 0.
или иначе 2sinαcosα =2sin(π/2)*cos(π/2) =2*1*0 = 0.
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B) cos(3π/2 -0,5x) =0,5 ;
-sin0,5x =0,5 ;
sin0,5x = - 0,5 ;
0,5x = (-1)^(n+1)*π/6 +π*n ,n∈Z.
x = (-1)^(n+1)*π/3 +2π*n ,n∈Z.
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C) только не sos =<span>2sinp/2sos2p=
</span>
6 -sin2π -cosπ +2sinπ/2cos2π =6 -0 -3*(-1)+2*0/2*1 =6+3=9.
