2х+5у=64
х+у=17
из второго
х=17-у
2(17-у)+5у=64
34-2у+5у=64
3у=30
у=10
х=17-10=7
В приведенном ниже решении замени все фигурные скобки на квадратные. Т.к. это совокупность (объединение) решений. Я не знаю как в латексе поставить квадратную скобку на несколько строк.
![x^2-|8,5x-16| \leq 8,5x\\ \left \{ {{x^2-8,5x+16 \leq 8,5x;8,5x-16 \geq 0 } \atop {x^2+8,5x-16 \leq 8,5x;8,5x-16<0 }} \right. \\\left \{ {{x^2-17x+16 \leq 0;x \geq 32/17 } \atop {x^2-16 \leq ;x<32/17 }} \right. \\\left \{ {{(x-1)(x-16)\leq 0;x \geq 32/17 } \atop {(x-4)(x+4) \leq ;x<32/17 }} \right. \\ \left \{ { 1\leq x \leq 16;x \geq 32/17 } \atop {-4\leq x \leq 4;x < 32/17 }} \right. \\ \left \{ {{32/17\leq x \leq 16} \atop {-4\leq x \leq 32/17}} \right. \\-4 \leq x \leq 16](https://tex.z-dn.net/?f=x%5E2-%7C8%2C5x-16%7C+%5Cleq+8%2C5x%5C%5C+%5Cleft+%5C%7B+%7B%7Bx%5E2-8%2C5x%2B16+%5Cleq+8%2C5x%3B8%2C5x-16+%5Cgeq+0+%7D+%5Catop+%7Bx%5E2%2B8%2C5x-16+%5Cleq+8%2C5x%3B8%2C5x-16%3C0+%7D%7D+%5Cright.+%5C%5C%5Cleft+%5C%7B+%7B%7Bx%5E2-17x%2B16+%5Cleq+0%3Bx+%5Cgeq+32%2F17+%7D+%5Catop+%7Bx%5E2-16+%5Cleq+%3Bx%3C32%2F17+%7D%7D+%5Cright.+%5C%5C%5Cleft+%5C%7B+%7B%7B%28x-1%29%28x-16%29%5Cleq+0%3Bx+%5Cgeq+32%2F17+%7D+%5Catop+%7B%28x-4%29%28x%2B4%29+%5Cleq+%3Bx%3C32%2F17+%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B+1%5Cleq+x+%5Cleq+16%3Bx+%5Cgeq+32%2F17+%7D+%5Catop+%7B-4%5Cleq+x+%5Cleq+4%3Bx+%3C+32%2F17+%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7B32%2F17%5Cleq+x+%5Cleq+16%7D+%5Catop+%7B-4%5Cleq+x+%5Cleq+32%2F17%7D%7D+%5Cright.+%5C%5C-4+%5Cleq+x+%5Cleq+16)
x ∈ [-4; 16]
длина промежутка = 20
D = 12*12+9*4*5 = 324
x1 = (12+18)/9*2 = 5/3 = 1 2/3
x2 = (12-18)/9*2 = -6/18 = -1/3
1) D(f)=R
2)f'(x)=(12x-x^3)=12-3x^2
3) 12-3x^2=0
3x^2-12=0
3x^2=12
x^2=12/3
x^2=4
x=4 f'(x)>0 4>0
Sin75=sin(30+45)=sin30*cos45+sin45*cos30= √2/2*(√3/2+1/2)=(√6+√2)/4
cos75=cos(45+30)=cos45*cos30 - sin45*sin30= √2/2(√3/2-1/2)= (√6-√2)/4