Question

ПРМОГИТЕ ПОЖАЛУЙСТА, СРОЧНО. ( первый раз спрашиваю, не знаю как работает)докажите что при любых значениях а правильное неравенство ( а+2 ) (а+4) >(а+1) (а+5)​

Answer 1

Ответ:

Объяснение:

$(a+2 ) (a+4) >(a+1) (a+5)\\( a+2)(a+4)=a^{2} +2a+4a+8=a^{2} +6a+8;\\(a+1)(a+5)=a^{2} +a+5a+5=a^{2} +6a+5\\a^{2} +6a+8>a^{2} +6a+5;\\8>5$

Answer 2

Proof of Inequality: (a+2)(a+4) > (a+1)(a+5)

To prove that the inequality (a+2)(a+4) > (a+1)(a+5) holds true for any value of 'a', we can expand both sides of the inequality and simplify the expression.

Expanding the left side of the inequality: (a+2)(a+4) = a(a+4) + 2(a+4) = a^2 + 4a + 2a + 8 = a^2 + 6a + 8

Expanding the right side of the inequality: (a+1)(a+5) = a(a+5) + 1(a+5) = a^2 + 5a + a + 5 = a^2 + 6a + 5

Now, let's compare the two expressions:

a^2 + 6a + 8 > a^2 + 6a + 5

Subtracting (a^2 + 6a) from both sides:

8 > 5

Since 8 is greater than 5, the inequality holds true for any value of 'a'. Therefore, we have proven that (a+2)(a+4) > (a+1)(a+5) for all values of 'a'.

Please let me know if you have any further questions!