Question

(y + 7)3 - y3 - 21y2 > 0​

Answer 1

Объяснение:

(y+7)³-y³-21y²>0

y³+21y²+147y+343-y³-21y²>0 (Сокращаем y³ и -y³, 21y² и -21y²).

147y+343>0

147y>-343

y>-343/147

y>7/3

$y = 2 \frac{2}{3}$

Answer 2

To solve the inequality $(y + 7)^3 - y^3 - 21y^2 > 0$, we can start by simplifying the expression:

$(y + 7)^3 - y^3 - 21y^2 > 0$

Expanding the cube of $y + 7$ using the binomial expansion, we have:

$(y^3 + 3y^2 \cdot 7 + 3y \cdot 7^2 + 7^3) - y^3 - 21y^2 > 0$

Simplifying further, we get:

$y^3 + 21y^2 + 147y + 343 - y^3 - 21y^2 > 0$

The $y^3$ and $-y^3$ terms cancel out, as well as the $21y^2$ and $-21y^2$ terms:

$147y + 343 > 0$

Subtracting 343 from both sides:

$147y > -343$

Finally, dividing both sides by 147 (remembering to reverse the inequality because we're dividing by a negative number), we obtain the solution:

$y < -\frac{343}{147}$

So, the solution to the inequality $(y + 7)^3 - y^3 - 21y^2 > 0$ is $y < -\frac{343}{147}$.