Решите уравнения: 1)-y^2+6=0; 2)3x^2-7=0; 3)-4x^2+19=0; 4)-0,3y^2+0,39=0; 5)3/2x^2-9/4=0

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Ответ:

luiza9505966348

06.12.2021

1)-y^2+6=0
-y^2=-6 умножаем на -1
y^2=6
у= корень из 6

2)3x^2-7=0
3x^2=7
x^2=7\3
x=корень из 7\3

3)-4x^2+19=0
-4x^2=-19 умножили на -1
4x^2=19
x^2=19\4
x = корень из 19\4

4)-0,3y^2+0,39=0
- 0.3y^2=-0.39 умножили на -1
0.3у^2= 0.39
у^2=0.39\0.3
у= корень из 0.39\0.3

5)3/2x^2-9/4=0
3/2x^2=9/4
x^2=1 (9/4 :3/2=2\2=1)
x= 1

4,4(50 оценок)

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Ответ:

misspotter

06.12.2021

Условие. Y²+xy-4x-9y+20=0 ; y=ax+1 ; x>2

найти все значения а, при которых графики имеют одну общую точку(в нашем случае (ax+1)² + x(ax+1) -4x - 9(ax+1)+20=0 имеет единственное решение).

Подставим у = (ax+1)² в уравнение у²+xy-4x-9y+20=0, получим

$(ax+1)^2+x(ax+1)-4x-9(ax+1)+20=0\\ a^2x^2+2ax+1+ax^2+x-4x-9ax-9+20=0\\ x^2(a^2+1)-(3+7a)x+12=0$

Найдем дискриминант квадратного уравнения относительно x

$D=(3+7a)^2-4(a^2+1)\cdot12=9+42a+49a^2-48a^2-48=\\ =a^2+42a-39=0$

Получим $a_{1,2}=-21\pm4\sqrt{30}$

Если подставить $a=-21+4\sqrt{30}$ , т.е. имеется квадратное уравнение $(922-168\sqrt{30})x^2+(144-28\sqrt{30})x+12=0$ , у которого корень

$\bigg(x-\dfrac{36+7\sqrt{30}}{29}\bigg)^2=0\\ \\ x=\dfrac{36+7\sqrt{30}}{29}2$

Если подставить $a=-21-4\sqrt{30}$ , т.е. имеется квадратное уравнение $(922+168\sqrt{30})x^2+(144+28\sqrt{30})x+12=0$ , у которого корень

$\bigg(x-\dfrac{36-7\sqrt{30}}{29}\bigg)^2=0\\ \\ x=\dfrac{36-7\sqrt{30}}{29}$

ответ: $a=-21+4\sqrt{30}$

4,6(78 оценок)

Ответ:

fhctybq123

06.12.2021

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4,4(31 оценок)

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