По неравенству Коши-Буняковского
![S^2 \leq (3^2+(-1)^2)(sin^2A+cos^2A) = 10 \cdot 1 = 10 \\ |S| \leq \sqrt{10}\\ S \leq \sqrt{10}](https://tex.z-dn.net/?f=S%5E2+%5Cleq+%283%5E2%2B%28-1%29%5E2%29%28sin%5E2A%2Bcos%5E2A%29+%3D+10+%5Ccdot+1+%3D+10+%5C%5C+%7CS%7C+%5Cleq+%5Csqrt%7B10%7D%5C%5C++S+%5Cleq+%5Csqrt%7B10%7D)
Ответ
9:Х=(9:2)-2
9:Х=2,5
2,5х=9
Х=9:2,5
Х=3,6
<span>1
x²y² - 6xy = -5
3x + 3y = 10⇒x+y=3 1/3
</span>
<span>x²y² - 6xy +5=0
xy=a
a²-6a+5=0
a1+a2=6 U a1+a2=5
a1=1 Ua2=5
1)xy=1
x+y=10/3⇒x=10/3-y
10/3y-y²=1
y²-10/3y+1=0
3y²-10y+3=0
D=100-36=64
y1=(10-8)/6=1/3⇒x1=10/3-1/3=3
y2=(10+8)/6=3⇒x2=10/3-3=1/3
2)xy=5
3y²-10y+5=0
D=100-60=40
y3=(10-2√10)/6=5/3-√10/3⇒x3=10/3-5/3+√10/3=5/3+√10/3
y4=5/3+√10/3⇒x4=10/3-5/3-√10/3=5/3-√10/3
(3;1/3);(1/3;3);((5+√10)/3;(5-√10)/3);((5-√10)/3;(5+√10/3)
2
x/y+5y/x=-6
2x+7y=6
</span>
<span><span>x/y+5y/x=-6</span>
</span>x/y=a
a+5/a=-6
a²+6a+5=0
a1+a2=-6 U a1*a2=5
a1=-5 U a2=-1
1)x/y=-5
x=-5y
-10y+7y=6
-3y=6
y=-2⇒x=10
2)x/y=-1
x=-y
-2y+7y=6
5y=6
y=1,2⇒x=-1,2
(10;-2);(-1,2;1,2)
(2a² +b)(a -2b²) =2a^3 -4a²b² +ab -2b^3