Question

Доказать неравенство 1) x(x+4)+6>4x 2) (a+5)(a-2)>(a-5)(a+8) 3) (a+6)²>12a

Answer 1

1)x(x+4)+6>4x
x²+4x+6>4x
x²+4x-4x>-6
x²>-6
Нет решения.
2)(a+5)(a-2)>(a-5)(a+8)
a²-2a+5a-10>a²+8a-5a-40
a²-2a+5a-a²-8a+5a>-40+10
0>-30
Неверно
Нет решения
3)(a+6)²>12a
a²+12a+36>12a
a²+12a-12a>-36
a²>-36
Нет решения

Answer 2

I will try to answer in detail and prove the inequalities.

1) x(x+4)+6>4x

To prove this inequality, we need to simplify it and find the values of x that make it true. We can do this by following these steps:

- Subtract 4x from both sides: x(x+4)+6-4x>0 - Expand the brackets: x^2+4x+6-4x>0 - Simplify: x^2>0 - Factor: x(x)>0 - Solve for x: x>0 or x<0

Therefore, the inequality is true for any x that is not equal to zero.

2) (a+5)(a-2)>(a-5)(a+8)

To prove this inequality, we need to simplify it and find the values of a that make it true. We can do this by following these steps:

- Expand the brackets: a^2+3a-10>a^2+3a-40 - Subtract a^2+3a from both sides: -10>-40 - Add 40 to both sides: 30>0

This is a true statement for any value of a, so the inequality is always true.

3) (a+6)²>12a

To prove this inequality, we need to simplify it and find the values of a that make it true. We can do this by following these steps:

- Expand the square: a^2+12a+36>12a - Subtract 12a from both sides: a^2+36>0 - Factor: (a+6)(a-6)>0 - Solve for a: a>-6 or a<6

Therefore, the inequality is true for any a that is not between -6 and 6.