The first item:
{ 5 x + 7 y + 2 = 0 4 x + 6 y + 3 = 0 \begin{cases}
5x + 7y + 2 = 0 \\
4x + 6y + 3 = 0
\end{cases} { 5 x + 7 y + 2 = 0 4 x + 6 y + 3 = 0
x = ∣ − 2 7 − 3 6 ∣ ∣ 5 7 4 6 ∣ = − 12 + 21 30 − 28 = 9 2 = 4.5 x = \frac{\begin{vmatrix}
-2 & 7\\
-3 & 6
\end{vmatrix}}
{\begin{vmatrix}
5 & 7\\
4 & 6
\end{vmatrix}} = \frac{-12 + 21}{30 - 28} = \frac{9}2 = 4.5 x = ∣ ∣ 5 4 7 6 ∣ ∣ ∣ ∣ − 2 − 3 7 6 ∣ ∣ = 30 − 28 − 12 + 21 = 2 9 = 4.5
y = ∣ 5 − 2 4 − 3 ∣ ∣ 5 7 4 6 ∣ = − 15 + 8 30 − 28 = − 7 2 = − 3.5 y = \frac{\begin{vmatrix}
5 & -2\\
4 & -3
\end{vmatrix}}
{\begin{vmatrix}
5 & 7\\
4 & 6
\end{vmatrix}} = \frac{-15 + 8}{30 - 28} = \frac{-7}2 = -3.5 y = ∣ ∣ 5 4 7 6 ∣ ∣ ∣ ∣ 5 4 − 2 − 3 ∣ ∣ = 30 − 28 − 15 + 8 = 2 − 7 = − 3.5
The second item:
1 4 + 2 6 + 3 8 + ⋯ = ∑ k = 1 ∞ k 2 k + 2 = ∑ k = 1 ∞ 1 2 k + 2 k ≥ ∑ k = 1 ∞ 1 4 k \frac{\sqrt{1}}{4} + \frac{\sqrt{2}}{6} + \frac{\sqrt{3}}{8} + \dots = \sum_{k=1}^{\infty} \frac{\sqrt{k}}{2k + 2} = \sum_{k=1}^{\infty} \frac{1}{2\sqrt{k} + \frac{2}{\sqrt{k}}} \ge \sum_{k=1}^{\infty} \frac{1}{4\sqrt{k}} 4 1 + 6 2 + 8 3 + ⋯ = ∑ k = 1 ∞ 2 k + 2 k = ∑ k = 1 ∞ 2 k + k 2 1 ≥ ∑ k = 1 ∞ 4 k 1
The last sum series is divergent, so the initial one is divergent too.
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