Question

Алгебра. Даю 30 балів.

Потрібно вирішити 3 завдання.

Answer 1

Відповідь:

1) 11

2) -130

3) tg²a

Пояснення:

1)

$\left \{ {{\sqrt{x} =4} \atop {x-2y=26}} \right.$

$\left \{ {{x=16} \atop {16-2y=26}} \right.$

-2y = 10

y = -5

x = 16, y = -5

Сума: 16-5 = 11

2)

$\left \{ {{y=3-x} \atop {x^2+4=8(3-x)}} \right.$

x² +4 - 24 +8x = 0

x² + 8x - 20 = 0

x₁ = -10

x₂ = 2

y₁ = 3 - (-10)

y₁ = 13

y₂ = 3-2

y₂ = 1

Перший добуток = -10 * 13 = -130

Другий добуток = 2 * 1 = 2

Менший добуток -130

3)

$(1+\frac{sin^2a}{cos^2a} )sin^2a$

Спільний множник:

$(\frac{cos^a+sin^2a}{cos^2a})sin^2a$

sin²a + cos²a = 1, (основна тригонометрична тотожність) звідси

$\frac{1}{cos^2a} * sin^2a = \frac{sin^2a}{cos2a} = tg^2a$

Answer 2

1.

$\displaystyle\bf\\\left \{ {{3 \sqrt{x} = 12 } \atop {x - 2y = 26 }} \right. \\ \\ 3 \sqrt{x} = 12 \\ \sqrt{x} = 12 \div 3 \\ \sqrt{x } = 4 \\ x = 16 \\ \\ 16 - 2y = 26 \\ 2y = 16 - 26 \\ 2y = - 10 \\ y = - 10 \div 2 \\ y = - 5 \\ \\ x + y = 16 - 5 = 11$

Ответ: 11

2.

$\displaystyle\bf\\\left \{ {{y + x = 3} \atop { {x}^{2} + 4 = 8y }} \right. \\ \displaystyle\bf\\\left \{ {{y = 3 - x} \atop { {x}^{2} + 4 = 8(3 - x) }} \right. \\ \\ {x}^{2} + 4 = 2 4- 8x \\ {x}^{2} + 8x - 20 = 0 \\ D = 8 {}^{2} - 4 \times ( - 20) = 64 + 80 = 144 \\ x_{1} = \frac{ - 8 - 12}{2} = - \frac{20}{2} = - 10\\ x_{2} = \frac{ - 8 + 12}{2} = \frac{4}{2} = 2 \\ \\ y_{1} =3 - ( - 10) = 3 + 10 = 13 \\ y_{2} = 3 - 2 = 1 \\ \\ x_{1}y_{1} = - 10 \times 13 = - 130 \\ x_{2}y_{2} = 2 \times 1 = 2$

Ответ: - 130

3.

$(1 + \tan {}^{2} ( \alpha ) ) \sin {}^{2} ( \alpha ) = (1 + \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } ) \sin {}^{2} ( \alpha ) = \\ = \frac{ \cos {}^{2} ( \alpha ) + \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } \times \sin {}^{2} ( \alpha ) = \frac{1}{ \cos {}^{2} ( \alpha ) } \times \sin {}^{2} ( \alpha ) = \\ = \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } = \tan {}^{2} ( \alpha )$

Ответ: 4) tg²a