Question

[Tú Trần]

tìm x \(\in\)Z biết |x+9| + |x+10| = 1

Answer 1

[Xyz OLM]

Ta có |x + 9| + |x + 10| = |x + 9| + |-x - 10| \(\ge\left|x+9-x-10\right|=\left|-1\right|=1\)

Dấu "=" xảy ra <=> (x + 9)(-x - 10) \(\ge0\)

TH1 : \(\hept{\begin{cases}x+9\ge0\\-x-10\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-9\\x\le-10\end{cases}}\Leftrightarrow x\in\varnothing\)

TH2 : \(\hept{\begin{cases}x+9\le0\\-x-10\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le-9\\x\ge-10\end{cases}}\Leftrightarrow-10\le x\le-9\)

Vậy \(-10\le x\le-9\)

Answer 2

[Đỗ Vũ Bá Linh]

Vì \(\hept{\begin{cases}\left|x+9\right|\ge0\\\left|x+10\right|\ge0\end{cases}}\) nên để \(\left|x+9\right|+\left|x+10\right|=1\) thì \(\orbr{\begin{cases}\left|x+9\right|=1\\\left|x+10\right|=1\end{cases}}\).

Ta có:

\(x+9\)	\(1\)	\(-1\)
\(x\)	\(-8\)	\(-10\)

\(x+10\)	\(1\)	\(-1\)
\(x\)	\(-9\)	\(-11\)

Vậy \(x\in\left\{-8;-9;-10;-11\right\}\).