Найдите сумму целых решений неравенства
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![x^2(x+3)^2-14x(x+3)+40 \leq 0\\ x(x+3)=t\\
t^2-14t+40 \leq 0\\ (t-4)(t-10) \leq 0
\\ \Rightarrow 4 \leq t \leq 10\ \Rightarrow 4 \leq x(x+3) \leq 10 \Leftrightarrow
\begin {cases} x^2+3x-10 \leq 0 \\ x^2+3x-4 \geq 0 \end {cases} \Leftrightarrow \\ \Leftrightarrow \begin {cases} (x+5)(x-2) \leq 0 \\ (x-1)(x+4)\geq 0 \end {cases} \Rightarrow \begin {cases} x \in [-5; 2] \\ x \in (-\infty; -4] \cup [1; +\infty) \end {cases} \Rightarrow \\ \\ \Rightarrow x \in [-5; -4] \cup [1;2]](https://tex.z-dn.net/?f=x%5E2%28x%2B3%29%5E2-14x%28x%2B3%29%2B40+%5Cleq+0%5C%5C+x%28x%2B3%29%3Dt%5C%5C%0At%5E2-14t%2B40+%5Cleq+0%5C%5C+%28t-4%29%28t-10%29+%5Cleq+0%0A%5C%5C+%5CRightarrow+4+%5Cleq+t+%5Cleq+10%5C+%5CRightarrow++4+%5Cleq+x%28x%2B3%29+%5Cleq+10+%5CLeftrightarrow++%0A%5Cbegin+%7Bcases%7D+x%5E2%2B3x-10+%5Cleq+0+%5C%5C+x%5E2%2B3x-4+%5Cgeq+0+%5Cend+%7Bcases%7D+%5CLeftrightarrow++%5C%5C+%5CLeftrightarrow++%5Cbegin+%7Bcases%7D+%28x%2B5%29%28x-2%29+%5Cleq+0+%5C%5C+%28x-1%29%28x%2B4%29%5Cgeq+0+%5Cend+%7Bcases%7D+%5CRightarrow+++%5Cbegin+%7Bcases%7D+x+%5Cin+%5B-5%3B+2%5D+%5C%5C+x+%5Cin+%28-%5Cinfty%3B+-4%5D+%5Ccup+%5B1%3B+%2B%5Cinfty%29+%5Cend+%7Bcases%7D+%5CRightarrow+%5C%5C+%5C%5C+%5CRightarrow+x+%5Cin+%5B-5%3B+-4%5D+%5Ccup+%5B1%3B2%5D "x^2(x+3)^2-14x(x+3)+40 \leq 0\\ x(x+3)=t\\
t^2-14t+40 \leq 0\\ (t-4)(t-10) \leq 0
\\ \Rightarrow 4 \leq t \leq 10\ \Rightarrow 4 \leq x(x+3) \leq 10 \Leftrightarrow
\begin {cases} x^2+3x-10 \leq 0 \\ x^2+3x-4 \geq 0 \end {cases} \Leftrightarrow \\ \Leftrightarrow \begin {cases} (x+5)(x-2) \leq 0 \\ (x-1)(x+4)\geq 0 \end {cases} \Rightarrow \begin {cases} x \in [-5; 2] \\ x \in (-\infty; -4] \cup [1; +\infty) \end {cases} \Rightarrow \\ \\ \Rightarrow x \in [-5; -4] \cup [1;2]")