![3^x*5^x=15^{3x}\\\\ (3*5)^{x}=15^{3x}\\\\ 15^x=15^{3x}\\\\ x=3x\\\\ 2x=0\\\\ x=0](https://tex.z-dn.net/?f=3%5Ex%2A5%5Ex%3D15%5E%7B3x%7D%5C%5C%5C%5C%0A%283%2A5%29%5E%7Bx%7D%3D15%5E%7B3x%7D%5C%5C%5C%5C%0A15%5Ex%3D15%5E%7B3x%7D%5C%5C%5C%5C%0Ax%3D3x%5C%5C%5C%5C%0A2x%3D0%5C%5C%5C%5C%0Ax%3D0)
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![\sqrt{x-3}=3 \sqrt[4]{x-3}+4\\\\ \sqrt[4]{x-3}=t \geq 0\\\\ t^2=3t+4\\\\ t^2-3t-4=0\\\\ t^2+t-4t-4=0\\\\ t(t+1)-4(t+1)=0\\\\ (t-4)(t+1)=0\\\\ t-4=0\ \ or\ \ t+1=0\\\\ t=4\ \ or\ \ t=-1\\\\ \sqrt[4]{x-3}=4\\\\ \left \{ {{x-3=4^4} \atop {4 \geq 0}} \right. \\\\ x=4^4+3=256+3=259\\\\](https://tex.z-dn.net/?f=+%5Csqrt%7Bx-3%7D%3D3+%5Csqrt%5B4%5D%7Bx-3%7D%2B4%5C%5C%5C%5C%0A++%5Csqrt%5B4%5D%7Bx-3%7D%3Dt+%5Cgeq+0%5C%5C%5C%5C%0At%5E2%3D3t%2B4%5C%5C%5C%5C%0At%5E2-3t-4%3D0%5C%5C%5C%5C%0At%5E2%2Bt-4t-4%3D0%5C%5C%5C%5C%0At%28t%2B1%29-4%28t%2B1%29%3D0%5C%5C%5C%5C%0A%28t-4%29%28t%2B1%29%3D0%5C%5C%5C%5C%0At-4%3D0%5C+%5C+or%5C+%5C+t%2B1%3D0%5C%5C%5C%5C%0At%3D4%5C+%5C+or%5C+%5C+t%3D-1%5C%5C%5C%5C%0A%5Csqrt%5B4%5D%7Bx-3%7D%3D4%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx-3%3D4%5E4%7D+%5Catop+%7B4+%5Cgeq+0%7D%7D+%5Cright.+%5C%5C%5C%5C%0Ax%3D4%5E4%2B3%3D256%2B3%3D259%5C%5C%5C%5C)
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![\sqrt{U}=\sqrt{T}\ \ \textless \ -\ \textgreater \ \ \left \{ {{U=T} \atop {T \geq 0}} \right. \\\\ \sqrt{x^2-2x}=\sqrt{x-2} \\\\ \left \{ {{x^2-2x=x-2} \atop {x-2 \geq0 }} \right. \\\\ \left \{ {{x^2-3x+2=0} \atop {x \geq 2}} \right. \\\\ \left \{ {{x^2-x-2x+2=0} \atop {x \geq 2}} \right. \\\\ \left \{ {{x(x-1)-2(x-1)=0} \atop {x \geq 2}} \right. \\\\ \left \{ {{(x-1)(x-2)=0} \atop {x \geq 2}} \right. \\\\ \left \{ {{x=1\ or\ x=2} \atop {x \geq 2}} \right. \\\\ x=2](https://tex.z-dn.net/?f=%5Csqrt%7BU%7D%3D%5Csqrt%7BT%7D%5C+%5C+%5Ctextless+%5C+-%5C+%5Ctextgreater+%5C+%5C+%5Cleft+%5C%7B+%7B%7BU%3DT%7D+%5Catop+%7BT+%5Cgeq+0%7D%7D+%5Cright.+%5C%5C%5C%5C%0A%0A+%5Csqrt%7Bx%5E2-2x%7D%3D%5Csqrt%7Bx-2%7D+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx%5E2-2x%3Dx-2%7D+%5Catop+%7Bx-2+%5Cgeq0+%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx%5E2-3x%2B2%3D0%7D+%5Catop+%7Bx+%5Cgeq+2%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx%5E2-x-2x%2B2%3D0%7D+%5Catop+%7Bx+%5Cgeq+2%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx%28x-1%29-2%28x-1%29%3D0%7D+%5Catop+%7Bx+%5Cgeq+2%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7B%28x-1%29%28x-2%29%3D0%7D+%5Catop+%7Bx+%5Cgeq+2%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx%3D1%5C+or%5C+x%3D2%7D+%5Catop+%7Bx+%5Cgeq+2%7D%7D+%5Cright.+%5C%5C%5C%5C%0Ax%3D2)
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![x^2-2x-3=x^2-2x+1-4=(x-1)^2-2^2=\\\\=[(x-1)-2]*[(x-1)+2]=(x-3)(x+1)](https://tex.z-dn.net/?f=x%5E2-2x-3%3Dx%5E2-2x%2B1-4%3D%28x-1%29%5E2-2%5E2%3D%5C%5C%5C%5C%3D%5B%28x-1%29-2%5D%2A%5B%28x-1%29%2B2%5D%3D%28x-3%29%28x%2B1%29)
![(0.4^{\frac{1}{x^2-2x-3}})^{6-x} \geq 1\\\\ 0.4^{\frac{6-x}{(x-3)(x+1)}} \geq 0.4^0\\\\ 0.4^{-\frac{(x-6)}{(x-3)(x+1)}} \geq 0.4^0\\\\ -\frac{(x-6)}{(x-3)(x+1)} \leq 0\\\\ \frac{x-6}{(x-3)(x+1)} \geq 0\\\\ -----(-1)++++++(3)-------[6]+++++\ \textgreater \ x\\\\ x\in(-1;\ 3)\cup[6;\ +\infty)](https://tex.z-dn.net/?f=%280.4%5E%7B%5Cfrac%7B1%7D%7Bx%5E2-2x-3%7D%7D%29%5E%7B6-x%7D+%5Cgeq+1%5C%5C%5C%5C%0A0.4%5E%7B%5Cfrac%7B6-x%7D%7B%28x-3%29%28x%2B1%29%7D%7D+%5Cgeq+0.4%5E0%5C%5C%5C%5C%0A0.4%5E%7B-%5Cfrac%7B%28x-6%29%7D%7B%28x-3%29%28x%2B1%29%7D%7D+%5Cgeq+0.4%5E0%5C%5C%5C%5C%0A-%5Cfrac%7B%28x-6%29%7D%7B%28x-3%29%28x%2B1%29%7D+%5Cleq+0%5C%5C%5C%5C%0A%5Cfrac%7Bx-6%7D%7B%28x-3%29%28x%2B1%29%7D+%5Cgeq+0%5C%5C%5C%5C%0A-----%28-1%29%2B%2B%2B%2B%2B%2B%283%29-------%5B6%5D%2B%2B%2B%2B%2B%5C+%5Ctextgreater+%5C+x%5C%5C%5C%5C%0Ax%5Cin%28-1%3B%5C+3%29%5Ccup%5B6%3B%5C+%2B%5Cinfty%29)
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![\sqrt{T}\ \textgreater \ \sqrt{U}\ \ \textless \ -\ \textgreater \ \ \left \{ {{T\ \textgreater \ U} \atop {U \geq 0}} \right. \\\\ \sqrt{3+7x}\ \textless \ \sqrt{1-4x} \\\\ \sqrt{1-4x}\ \textgreater \ \sqrt{3+7x}\\\\ \left \{ {{1-4x\ \textgreater \ 3+7x} \atop {3+7x \geq 0}} \right. \\\\ \left \{ {{-11x\ \textgreater \ 2} \atop {7x \geq -3}} \right. \\\\ \left \{ {{x\ \textless \ -\frac{2}{11}} \atop {x \geq -\frac{3}{7}}} \right. \\\\ -\frac{3}{7}\leq x\ \textless \ -\frac{2}{11}\\\\ x\in[-\frac{3}{7};\ -\frac{2}{11})](https://tex.z-dn.net/?f=%5Csqrt%7BT%7D%5C+%5Ctextgreater+%5C+%5Csqrt%7BU%7D%5C+%5C+%5Ctextless+%5C+-%5C+%5Ctextgreater+%5C+%5C+%5Cleft+%5C%7B+%7B%7BT%5C+%5Ctextgreater+%5C+U%7D+%5Catop+%7BU+%5Cgeq+0%7D%7D+%5Cright.++%5C%5C%5C%5C%0A%0A+%5Csqrt%7B3%2B7x%7D%5C+%5Ctextless+%5C+%5Csqrt%7B1-4x%7D+%5C%5C%5C%5C%0A+%5Csqrt%7B1-4x%7D%5C+%5Ctextgreater+%5C++%5Csqrt%7B3%2B7x%7D%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7B1-4x%5C+%5Ctextgreater+%5C+3%2B7x%7D+%5Catop+%7B3%2B7x+%5Cgeq+0%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7B-11x%5C+%5Ctextgreater+%5C+2%7D+%5Catop+%7B7x+%5Cgeq+-3%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+-%5Cfrac%7B2%7D%7B11%7D%7D+%5Catop+%7Bx+%5Cgeq+-%5Cfrac%7B3%7D%7B7%7D%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+-%5Cfrac%7B3%7D%7B7%7D%5Cleq+x%5C+%5Ctextless+%5C+-%5Cfrac%7B2%7D%7B11%7D%5C%5C%5C%5C%0Ax%5Cin%5B-%5Cfrac%7B3%7D%7B7%7D%3B%5C+-%5Cfrac%7B2%7D%7B11%7D%29)
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![3*cos(\frac{3\pi}{2}+x)-5*cos(x)=0\\\\ 3*sin(x)-5*cos(x)=0\\\\](https://tex.z-dn.net/?f=3%2Acos%28%5Cfrac%7B3%5Cpi%7D%7B2%7D%2Bx%29-5%2Acos%28x%29%3D0%5C%5C%5C%5C%0A3%2Asin%28x%29-5%2Acos%28x%29%3D0%5C%5C%5C%5C)
Если в уравнении положить
![cos(x)=0](https://tex.z-dn.net/?f=cos%28x%29%3D0)
, то из уравнения следует, что и
![sin(x)=0](https://tex.z-dn.net/?f=sin%28x%29%3D0)
, чего быть не может (синус и косинус того же аргумента не могут равняться нулю одновременно, это противоречит основному тригонометрическому тождеству), т.е. в данном уравнении
![cos(x) \neq 0](https://tex.z-dn.net/?f=cos%28x%29+%5Cneq+0)
Это означает, что мы можем делить уравнение на
![cos(x)](https://tex.z-dn.net/?f=cos%28x%29)
и решение нового уравнения совпадать с решениям исходного (равносильный переход)
![3*sin(x)-5*cos(x)=0\\\\ 3*sin(x)=5*cos(x)\\\\ 3*\frac{sin(x)}{cos(x)}=5*\frac{cos(x)}{cos(x)}\\\\ 3*tg(x)=5\\\\ tg(x)=\frac{5}{3}\\\\ x=arctg(\frac{5}{3})+\pi n,\ \ n\in Z](https://tex.z-dn.net/?f=3%2Asin%28x%29-5%2Acos%28x%29%3D0%5C%5C%5C%5C%0A3%2Asin%28x%29%3D5%2Acos%28x%29%5C%5C%5C%5C%0A3%2A%5Cfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%3D5%2A%5Cfrac%7Bcos%28x%29%7D%7Bcos%28x%29%7D%5C%5C%5C%5C%0A3%2Atg%28x%29%3D5%5C%5C%5C%5C%0Atg%28x%29%3D%5Cfrac%7B5%7D%7B3%7D%5C%5C%5C%5C%0Ax%3Darctg%28%5Cfrac%7B5%7D%7B3%7D%29%2B%5Cpi+n%2C%5C+%5C+n%5Cin+Z)
ОДЗ
x ≥ 0,5
3x^2 - x - 2 = 0
D = 1 + 24 = 25
x1 = ( 1 + 5)/6 = 1;
x2 = ( 1 - 5)/6 = - 2/3 ==> не удовлет ОДЗ
2x - 1 = 0
x = 0,5
Ответ
1; 0,5
Ответ:
1
Объяснение:
т.к. мы берём все числа степени в диапазоне от 1 до 1000.
1000-1+1=1000
1000 это четное число, а любое число в четной степени положительно
Эту систему будем решать методом подстановки.
х² - 3ху + 2у² = 3 |* (-2) -2х² +6ху - 4у² = -6
2х² - 2ху - у² = - 6 , ⇒ <span>2х² - 2ху - у² = - 6 </span> Сложим
получаем:
4ху - 5у² = -12
4ху = 5у² -12
х = (5у² -12)/4у это и есть наша подстановка. Подставим в любое из данных уравнений, например, в 1-е.
( (5у² -12)/4у)² - 3у*(5у² - 12)/4у + 2у² = 3
(25у⁴ - 120у² + 144)/16у² - (15у² -36)/4 + 2у² = 3 | * 16у²
25у⁴ - 120у² + 144 - 4у²(15у² -36) + 32у⁴ = 48у²
25у⁴ - 120у² + 144 - 60у⁴ + 144у² + 32у⁴ -48у² = 0
-3у⁴-24у² + 144 = 0
у⁴ + 8у² - 48 = 0
Это биквадратное уравнение. у² = t
t² + 8t -48 = 0 По т. Виета t₁ = -12, t₂ = 4
a) t₁ = -12
y² = -12
нет решений
б) t₂ = 4
y² = 4
y₁ = 2, y₂ = -2
Теперь ищем х
1) у₁ = 2
х₁ = (5у² -12)/4у=(20 -12)/8 = 1 решение: (1; 2)
2) у₂ = -2
х₂ - (20 -12)/(-8) = -1 решение (-1;-2)
Для каждой пары ищем х₀ + х₀у₀ + у₀
<span>(1; 2)
</span><span>х₀ + х₀у₀ + у₀ = 1 + 2 + 2 = 5
</span>(-1;-2)
<span>х₀ + х₀у₀ + у₀ = -1 +2 -2 = -1
</span>Ответ: -1