Question

Помогите пожалуйста!!!!!

Answer 1

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Answer 2

Уравнения:
1)
$frac{2}{5} *x^2 = frac{1}{3} \ \ frac{2x^2}{5} = frac{1}{3} \ \ 3*2x^2 = 1*5 \ \ 6x^2 =5 \ \ x^2= frac{5}{6} \ \ x= sqrt{ frac{5}{6} } \ \ x_{1} = sqrt{ frac{5}{6} } = frac{ sqrt{30} }{6} \ \ x_{2} = - sqrt{ frac{5}{6} } = - frac{ sqrt{30} }{6}$

2) 
$frac{4}{x-2} = frac{3}{4-x^2}$
знаменатели ≠0
х-2≠0  ⇒  х≠2
4-х²≠0  ;  х≠√4  ⇒ х≠2  ; х≠-2
4(4-х²) = 3(х-2)
16 -4х² = 3х -6
16 -4х² -3х +6 =0
-4х² -3х +22=0
D= (-3)² - 4*(-4)*22= 9 + 352=361=19²
x₁= (3 -19)/ (2*(-4)) = -16/ (-8)= 2
x₂= (3+19)/ (-8) = 22/ (-8) = -  11/4= - 2.75

Упростить:
$frac{2}{2- sqrt{7}} + frac{3}{2+ sqrt{7} } + frac{ sqrt{7} }{3} = \ \ = frac{2*3*(2+ sqrt{7}) + 3*3*(2- sqrt{7})+ sqrt{7}(2- sqrt{7})(2+ sqrt{7} ) }{3(2- sqrt{7})(2+ sqrt{7}) } = \ \ = frac{12+6 sqrt{7}+18-9 sqrt{7} + sqrt{7} (2^2-( sqrt{7})^2) }{3(2^2- ( sqrt{7})^2) } = \ \ = frac{30 -3 sqrt{7} -3 sqrt{7} }{3 (4-7)} = frac{30 -6 sqrt{7} }{-9} = \ \ = frac{10-2 sqrt{7} }{-3} =frac{2}{3} (sqrt{7} -5)$

Система:
{x+3y= 10
{xy=3

{x=10-3y
{xy=3

y(10-3y)=3
10y -3y² -3 =0     |*(-1)
3y² -10y +3 =0
D= (-10)² -4*3*3 = 100 -36=64=8²
y₁= (10-8)/(2*3) = 2/6= 1/3
y₂= (10+8)/6 =18/6 = 3

x₁= 10 - 3 * 1/3 = 10-1=9
x₂= 10 - 3*3 = 1