Методом вычитания
2x+3y=-1 |*5
5x+4y=1 |*2
10x+15y=-5
10x+8y=2
__________
7y=-7
y=-1
подставляем в любое уравнение:
2x+3*(-1)=-1
2x-3=-1
2x=-1+3
2x=2
x=1
проверка:
5x+4y=1
5*1+4*(-1)=1
5-4=1
1=1
ответ: x=1, y=-1
![1)\ 5(sinx+cosx)+sin2x+1=0\\ 5(sinx+cosx)+(2sinxcosx+sin^2x+cos^2x)=0\\ 5(sinx+cosx)+(sinx+cosx)^2=0\\ (sinx+cosx)(5+sinx+cosx)=0\\ 1.\ sinx+cosx=0\\ if\ cosx=0\ => \ sinx+0=0\ => sinx=0\\ sin^2x+cos^2x=1\\ 0+0=1\ =>\ cosx\neq 0\\ \frac{sinx}{cosx}+1=0\\ tgx=-1\\ x=\frac{3\pi}{4}+\pi k\\ 2.\ 5+sinx+cosx=0\\ sinx+cosx=-5\\ -1 \leq sinx \leq 1\\ -1 \leq cosx \leq 1\\ -2 \leq sinx+cosx \leq 2\\ -5<-2](https://tex.z-dn.net/?f=1%29%5C+5%28sinx%2Bcosx%29%2Bsin2x%2B1%3D0%5C%5C%0A5%28sinx%2Bcosx%29%2B%282sinxcosx%2Bsin%5E2x%2Bcos%5E2x%29%3D0%5C%5C%0A5%28sinx%2Bcosx%29%2B%28sinx%2Bcosx%29%5E2%3D0%5C%5C%0A%28sinx%2Bcosx%29%285%2Bsinx%2Bcosx%29%3D0%5C%5C%0A1.%5C+sinx%2Bcosx%3D0%5C%5C%0Aif%5C+cosx%3D0%5C+%3D%3E+%5C+sinx%2B0%3D0%5C+%3D%3E+sinx%3D0%5C%5C%0Asin%5E2x%2Bcos%5E2x%3D1%5C%5C%0A0%2B0%3D1%5C+%3D%3E%5C+cosx%5Cneq+0%5C%5C%0A%5Cfrac%7Bsinx%7D%7Bcosx%7D%2B1%3D0%5C%5C%0Atgx%3D-1%5C%5C%0Ax%3D%5Cfrac%7B3%5Cpi%7D%7B4%7D%2B%5Cpi+k%5C%5C%0A2.%5C+5%2Bsinx%2Bcosx%3D0%5C%5C%0Asinx%2Bcosx%3D-5%5C%5C%0A-1+%5Cleq+sinx+%5Cleq+1%5C%5C%0A-1+%5Cleq+cosx+%5Cleq+1%5C%5C%0A-2+%5Cleq+sinx%2Bcosx+%5Cleq+2%5C%5C%0A-5%3C-2)
Решений нет.
Ответ:
![\frac{3\pi}{4}+\pi k](https://tex.z-dn.net/?f=%5Cfrac%7B3%5Cpi%7D%7B4%7D%2B%5Cpi+k)
; k - целое
![2) sin12x-sin4x=2\\ -1 \leq sin12x \leq 1\\ -1 \leq sin4x \leq 1\ => \ -1 \leq -sin4x \leq 1\\ -2 \leq sin12x-sin4x \leq 2 \\ sin12x-sin4x=2\ <=>\ \left \{ {{sin12x=1} \atop {-sin4x=1}} \right. \\ \left \{ {{12x=\frac{\pi}{2}+2\pi k} \atop {sin4x=-1}} \right. \\ \left \{ {{x=\frac{\pi}{24}+\frac{\pi}{6} k} \atop {4x=\frac{3\pi}{2}+2\pi k}} \right. \\ \left \{ {{x=\frac{\pi}{24}+\frac{\pi}{6} k} \atop {x=\frac{3\pi}{8}+\frac{\pi}{2} k}} \right. \\](https://tex.z-dn.net/?f=2%29+sin12x-sin4x%3D2%5C%5C%0A-1+%5Cleq+sin12x+%5Cleq+1%5C%5C%0A-1+%5Cleq+sin4x+%5Cleq+1%5C+%3D%3E+%5C+-1+%5Cleq+-sin4x+%5Cleq+1%5C%5C%0A-2+%5Cleq+sin12x-sin4x+%5Cleq+2+%5C%5C%0Asin12x-sin4x%3D2%5C+%3C%3D%3E%5C++%5Cleft+%5C%7B+%7B%7Bsin12x%3D1%7D+%5Catop+%7B-sin4x%3D1%7D%7D+%5Cright.+%5C%5C%0A+%5Cleft+%5C%7B+%7B%7B12x%3D%5Cfrac%7B%5Cpi%7D%7B2%7D%2B2%5Cpi+k%7D+%5Catop+%7Bsin4x%3D-1%7D%7D+%5Cright.+%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx%3D%5Cfrac%7B%5Cpi%7D%7B24%7D%2B%5Cfrac%7B%5Cpi%7D%7B6%7D+k%7D+%5Catop+%7B4x%3D%5Cfrac%7B3%5Cpi%7D%7B2%7D%2B2%5Cpi+k%7D%7D+%5Cright.+%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx%3D%5Cfrac%7B%5Cpi%7D%7B24%7D%2B%5Cfrac%7B%5Cpi%7D%7B6%7D+k%7D+%5Catop+%7Bx%3D%5Cfrac%7B3%5Cpi%7D%7B8%7D%2B%5Cfrac%7B%5Cpi%7D%7B2%7D+k%7D%7D+%5Cright.+%5C%5C)
Так как это система, то точки должны совпадать. Если изобразить эти решения на единичной окружности, становится понятно, что 2-е решение является "подрешением" 1-го, а потому именно 2-е является решением всей системы.
Ответ:
![\frac{3\pi}{8}+\frac{\pi}{2}k](https://tex.z-dn.net/?f=%5Cfrac%7B3%5Cpi%7D%7B8%7D%2B%5Cfrac%7B%5Cpi%7D%7B2%7Dk)
; k - целое
![3)\ (x^2-\frac{4\pi}{3}x+\frac{\pi^2}{3})*arccosx=0\\ -1 \leq x \leq 1\\ 1. \ x^2-\frac{4\pi}{3}x+\frac{\pi^2}{3}=0\\ D_1=\frac{4\pi^2}{9}-\frac{\pi^2}{3}=\frac{4\pi^2-3\pi^2}{9}=\frac{\pi^2}{9}\\ |x=\frac{2\pi}{3}+\frac{\pi}{3}\\ |x=\frac{2\pi}{3}-\frac{\pi}{3}\\ \\ |x=\pi;\ -1\leq x \leq 1;\ \pi>1.\\ |x=\frac{\pi}{3};\ \pi>3\ =>\ \frac{\pi}{3}>1.\\ 2.\ arccosx=0\\ x=0\\ ](https://tex.z-dn.net/?f=3%29%5C+%28x%5E2-%5Cfrac%7B4%5Cpi%7D%7B3%7Dx%2B%5Cfrac%7B%5Cpi%5E2%7D%7B3%7D%29%2Aarccosx%3D0%5C%5C%0A-1+%5Cleq+x+%5Cleq+1%5C%5C%0A1.+%5C+x%5E2-%5Cfrac%7B4%5Cpi%7D%7B3%7Dx%2B%5Cfrac%7B%5Cpi%5E2%7D%7B3%7D%3D0%5C%5C%0AD_1%3D%5Cfrac%7B4%5Cpi%5E2%7D%7B9%7D-%5Cfrac%7B%5Cpi%5E2%7D%7B3%7D%3D%5Cfrac%7B4%5Cpi%5E2-3%5Cpi%5E2%7D%7B9%7D%3D%5Cfrac%7B%5Cpi%5E2%7D%7B9%7D%5C%5C%0A%7Cx%3D%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B%5Cfrac%7B%5Cpi%7D%7B3%7D%5C%5C%0A%7Cx%3D%5Cfrac%7B2%5Cpi%7D%7B3%7D-%5Cfrac%7B%5Cpi%7D%7B3%7D%5C%5C%0A%5C%5C%0A%7Cx%3D%5Cpi%3B%5C+-1%5Cleq+x+%5Cleq+1%3B%5C+%5Cpi%3E1.%5C%5C%0A%7Cx%3D%5Cfrac%7B%5Cpi%7D%7B3%7D%3B%5C+%5Cpi%3E3%5C+%3D%3E%5C+%5Cfrac%7B%5Cpi%7D%7B3%7D%3E1.%5C%5C%0A2.%5C+arccosx%3D0%5C%5C%0Ax%3D0%5C%5C%0A)
Ответ: 0
Решение:
28y^2/ ∛7y
Чтобы избавиться от иррациональности в знаменателе умножим числитель и знаменатель ∛(7y)^2
28y^2*∛(7y)^2 / ∛7y*∛(7y)^2=4^1*7^1*y^2*(7y)^2/3 : ∛(7y)^3=4^1*7^1*y^2*7^2/3*y^2/3 :7у=4^1*7^1*y^2*7^2/3*y^2/3*7^-1*y^-1=4*7^(1+2/3-1)*y^(2+2/3-1)=4*7^2/3*y^(1+2/3)=4*7^2/3*y*y^2/3=4y*∛(7^2*y^2)=4y*∛49y^2
Ответ: 4y∛49y^2
Под корнем 475,24-331,24= под корнем 144= 12
под корнем 289-4096= под корнем -3807= под корнем 9*(-423)= 3 под корнем -423= 3 под корнем 9*(-47)= 9 <span>под корнем-47
</span>под корнем 13689-11664= <span>под корнем 2025= 45</span>