Giải bất phương trình |x + 1| + |x - 4| > 7. Giá trị nghiệm nguyên dương nhỏ nhất của x thoả bất

Chọn đáp án D

Xét dấu phá trị tuyệt đối:

TH1. $x\in \left(-\infty ;-1\right)$

$\left|x+1\right|+\left|x-4\right|>7\iff \left\{\begin{array}{l}x\in \left(-\infty ;-1\right)\\ -\left(x+1\right)-\left(x-4\right)>7\end{array}\right.$

$\iff \left\{\begin{array}{l}x\in \left(-\infty ;-1\right)\\ -2x+3>7\end{array}\right.\iff \left\{\begin{array}{l}x\in \left(-\infty ;-1\right)\\ x<-2\end{array}\right.\iff x\in \left(-\infty ;-2\right)$

TH2. $x\in \left[-1;4\right)$

$\left|x+1\right|+\left|x-4\right|>7\iff \left\{\begin{array}{l}x\in \left[-1;4\right)\\ \left(x+1\right)-\left(x-4\right)>7\end{array}\right.\iff \left\{\begin{array}{l}x\in \left[-1;4\right)\\ 5>7\end{array}\right.\iff x\in \varnothing$

TH3. $x\in \left[4;+\infty \right)$

$\left|x+1\right|+\left|x-4\right|>7\iff \left\{\begin{array}{l}x\in \left[4;+\infty \right)\\ \left(x+1\right)+\left(x-4\right)>7\end{array}\right.\iff \left\{\begin{array}{l}x\in \left[4;+\infty \right)\\ 2x-3>7\end{array}\right.$

$\iff \left\{\begin{array}{l}x\in \left[4;+\infty \right)\\ x>5\end{array}\right.\iff x\in \left(5;+\infty \right)$