<span>{4x+y=14
{x−y=10
Складываем эти уравнения
4х + у + х - у = 14 + 10
5х = 24
х = 24 : 5
х = 4,8
Во второе уравнение х - у = 10 подставим х = 4,8 и найдём у.
4,8 - у = 10
- у = 10 - 4,8
- у = 5,2
у = 5,2 : (-1)
у = - 5,2
Проверка x = 4,8; y = - 5,2
</span>{4 · 4,8 + (-5,2) =14 => 19,2 - 5,2 = 14 => 14=14
{4,8 − (-5,2) =10 => 1,8 + 5,2 = 10 => 10=10
Ответ: x = 4,8; y = - 5,2.
Вроде бы так, только, думаю, у тебя она ровней получится)
Выражения под корнями взаимно обратные. Сделаем замену :
![(\sqrt{3-2\sqrt{2} })^{x}=m,m>0](https://tex.z-dn.net/?f=%28%5Csqrt%7B3-2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%3Dm%2Cm%3E0)
Тогда :
![(\sqrt{3+2\sqrt{2} })^{x}=\frac{1}{m}](https://tex.z-dn.net/?f=%28%5Csqrt%7B3%2B2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%3D%5Cfrac%7B1%7D%7Bm%7D)
![m+\frac{1}{m} \geq 6\\\\m^{2}-6m+1\geq0\\\\m^{2}-6m+1=0\\\\D=36-4=32=4\sqrt{2}\\\\m_{1}=\frac{6-4\sqrt{2} }{2}=3-2\sqrt{2}\\\\m_{2}=\frac{6+4\sqrt{2} }{2}=3+2\sqrt{2}](https://tex.z-dn.net/?f=m%2B%5Cfrac%7B1%7D%7Bm%7D%20%5Cgeq%206%5C%5C%5C%5Cm%5E%7B2%7D-6m%2B1%5Cgeq0%5C%5C%5C%5Cm%5E%7B2%7D-6m%2B1%3D0%5C%5C%5C%5CD%3D36-4%3D32%3D4%5Csqrt%7B2%7D%5C%5C%5C%5Cm_%7B1%7D%3D%5Cfrac%7B6-4%5Csqrt%7B2%7D%20%7D%7B2%7D%3D3-2%5Csqrt%7B2%7D%5C%5C%5C%5Cm_%7B2%7D%3D%5Cfrac%7B6%2B4%5Csqrt%7B2%7D%20%7D%7B2%7D%3D3%2B2%5Csqrt%7B2%7D)
![(m-(3-\sqrt{2}))(m-(3+2\sqrt{2}))\geq 0](https://tex.z-dn.net/?f=%28m-%283-%5Csqrt%7B2%7D%29%29%28m-%283%2B2%5Csqrt%7B2%7D%29%29%5Cgeq%200)
+ - +
0_________[3-2√2]__________[3 + 2√2]__________ m
1) 0 < m ≤ 3 - 2√2 2) m ≥ 3 + 2√2
![1)(\sqrt{3-2\sqrt{2} })^{x} \leq3-2\sqrt{2}\\\\(3-2\sqrt{2})^{\frac{x}{2} }\leq 3-2\sqrt{2}\\\\\frac{x}{2} \leq1\\\\x\leq2\\\\x\in(-\infty;2]\\\\2)(\sqrt{3-2\sqrt{2} })^{x}\geq3+2\sqrt{2}\\\\(3-2\sqrt{2})^{\frac{x}{2} }\geq (3-2\sqrt{2})^{-1}\\\\\frac{x}{2}\geq-1\\\\x\geq -2\\\\x\in[-2;+\infty)](https://tex.z-dn.net/?f=1%29%28%5Csqrt%7B3-2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%20%5Cleq3-2%5Csqrt%7B2%7D%5C%5C%5C%5C%283-2%5Csqrt%7B2%7D%29%5E%7B%5Cfrac%7Bx%7D%7B2%7D%20%7D%5Cleq%203-2%5Csqrt%7B2%7D%5C%5C%5C%5C%5Cfrac%7Bx%7D%7B2%7D%20%5Cleq1%5C%5C%5C%5Cx%5Cleq2%5C%5C%5C%5Cx%5Cin%28-%5Cinfty%3B2%5D%5C%5C%5C%5C2%29%28%5Csqrt%7B3-2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%5Cgeq3%2B2%5Csqrt%7B2%7D%5C%5C%5C%5C%283-2%5Csqrt%7B2%7D%29%5E%7B%5Cfrac%7Bx%7D%7B2%7D%20%7D%5Cgeq%20%283-2%5Csqrt%7B2%7D%29%5E%7B-1%7D%5C%5C%5C%5C%5Cfrac%7Bx%7D%7B2%7D%5Cgeq-1%5C%5C%5C%5Cx%5Cgeq%20-2%5C%5C%5C%5Cx%5Cin%5B-2%3B%2B%5Cinfty%29)
Ответ : x ∈ [- 2 ; 2]
1
sin^4a+cos^4a=(1-cos2a)²/4+(1+cos2a)²/4=
=1/4*(1-2cos2a+cos²2a+1+2cos2a+cos²2a)=1/4*(2+2cos²2a)=
=1/2*(1+cos²2a)
2
ctga-tga=cosa/sina-sina/cosa=(cos²a-sin²a(/(sinacosa)=cos2a:1/2*sin2a=
=2cos2a/sin2a=2ctg2a
3
(1-tg²a)/(1+tg²a)=(1-sin²a/cos²a):(<span>1+sin²a/cos²a)=
=(cos</span>²a-sin²a)/cos²a:(cos²a+sin²a)/cos²a=cos2a/cos²a*cos²a/1=cos2a