Помогите пожалуйста!!!!!. Created by olaola350. algebra-ru

Уравнения:1)2) знаменатели ≠0х-2≠0 ⇒ х≠24-х²≠0 ; х≠√4 ⇒ х≠2 ; х≠-24(4-х²) = 3(х-2)16 -4х² = 3х -616 -4х² -3х +6 =0-4х² -3х +22=0D= (-3)² - 4*(-4)*22= 9 + 352=361=19²x₁= (3 -19)/ (2*(-4)) = -16/ (-8)= 2x₂= (3+19)/ (-8) = 22/ (-8) = - 11/4= - 2.75Упростить:Система:{x+3y= 10{xy=3{x=10-3y{xy=3y(10-3y)=310y -3y² -3 =0 |*(-1)3y² -10y +3 =0D= (-10)² -4*3*3 = 100 -36=64=8²y₁= (10-8)/(2*3) = 2/6= 1/3y₂= (10+8)/6 =18/6 = 3x₁= 10 - 3 * 1/3 = 10-1=9x₂= 10 - 3*3 = 1Aigul9912Aigul9912======================== Новые вопросы в АлгебраРеши уравнение. Помогите пожалуйстаРеши систему уравнений .Найти область определения функциидам 20 баллов!!Найти область определения функции 20 б.Найди область определения функции 20 балловнайди область определения функции20 БАЛЛОВСрочно помогите пожалуйста!Представьте в виде произведения выражение 100-4а^2 и 1000m+m^4СРОЧНО, ДАЮ 40 БАЛЛОВ{x+ay=11{ax+y=-1{3y-2x=11Неотрицательные числа x и y удовлетворяют неравенству xy >= x + 4y + 2 . 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Помогите пожалуйста!!!!!...

olaola350 @olaola350

July 2022 1 39 Report

Помогите пожалуйста!!!!!

Answers & Comments

Aigul9912 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