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]
<span>Планировалось скосить луг за x дней, скосили за x-2 дня, площадь луга 20*x или (x-2)*25
</span><span>Таким образом, составим уравнение:
</span>20*x=(x-2)*25
20x=25x-50
20x-25x=-50
-5x=-50
x=-50:(-5)
x=10
Тоесть планировалось скосить за 10 дней.
S=20*10=200 га²
Ответ: 200 га²
ctg=1/tg поэтому, ctg=4/корень из 3!
{x+8≥0⇒x≥-8
{x-4≥0⇒x≥4
x∈[4;∞)
(x-4)²=(√x+8)²
x²-8x+16=x+8
x²-9x+8=0
x1+x2=9 U x1*x2=8
x1=1 не удов усл
х2=8
(x-5)^2-x+3=0; x^2-11x+28=0; D=121-4*28=9; x1=11-3/2=4;x2=11+3/2=7