Tìm x biết |x-1 |+ |x-4 |=3x

\(\dfrac{a^2+a+3}{a+1}\in Z\)

\(\Rightarrow a^2+a+3⋮a+1\)

\(\Rightarrow a^2+a-2a-2+5⋮a+1\)

\(\Rightarrow a\left(a+1\right)-2\left(a-1\right)+5⋮a+1\)

\(\Rightarrow\left(a-2\right)\left(a-1\right)+5⋮a+1\)

\(\Rightarrow5⋮a+1\)

\(\Rightarrow a+1\inƯ\left(5\right)\)

\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}a+1=1\Rightarrow a=0\\a+1=-1\Rightarrow a=2\\a+1=5\Rightarrow a=4\\a+1=-5\Rightarrow a=-6\end{matrix}\right.\)

2)\(\left|x-1\right|+\left|x-4\right|=3x\)

\(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\\left|x-4\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x-1\right|+\left|x-4\right|\ge0\Rightarrow3x\ge0\)

\(\Rightarrow x-1+x-4=3x\)

\(\Rightarrow2x-5=3x\Rightarrow x=-\dfrac{2}{5}\)

tương tự

3)\(A=\left|x+7\right|+\left|x-5\right|+5\)

\(A=\left|x+7\right|+\left|5-x\right|+5\)

Áp dụng bđt:

\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

\(\Rightarrow A\ge\left|x+7+5-x\right|\)

\(\Rightarrow A\ge17\)

Dấu "=" xảy ra khi;

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+7\ge0\Rightarrow x\ge-7\\5-x\ge0\Rightarrow x\le5\end{matrix}\right.\\\left\{{}\begin{matrix}x+7< 0\Rightarrow x< 7\\5-x< 0\Rightarrow x>5\end{matrix}\right.\end{matrix}\right.\)

Vậy xảy ra khi

\(-7\le x\le5\)

tương tự mấy câu dễ tự làm