Помогите пожалуйста!!!!!

Question 1

Question 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

Question 3

========================

zhenyaM2002_zn · Answer 1 · 2018-04-29T19:13:02+0000

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