Question

/2x+3/ + /2x-1/ =\(\dfrac{8}{2\left(y-5\right)^2+2}\)

Đề có hơi nâng cao nha !

 

Answer 1

Ta thấy \(\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge4\)

Và \(\dfrac{8}{2\left(y-5\right)^2+2}\le\dfrac{8}{2\cdot0+2}=4\)

Do đó \(VT\ge4\ge VP\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3\right)\left(1-2x\right)\ge0\\y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{2}\le x\le\dfrac{1}{2}\\y=5\end{matrix}\right.\)

Vậy PT có nghiệm x trong khoảng \(\left[-\dfrac{3}{2};\dfrac{1}{2}\right]\) và \(y=5\)