Question

1,tìm x biết:\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3.\left(x+1\right)^2+2}\)

Answer 1

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

\(VP=\frac{8}{3\left(x+1\right)^2+2}\le\frac{8}{2}=4\)

\(VT\ge VP\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(2x+3\right)\left(1-2x\right)\ge0\left(1\right)\\\left(x+1\right)^2=0\left(2\right)\end{cases}}\)

\(\left(2\right)\)\(\Leftrightarrow\)\(x=-1\) ( thỏa mãn\(\left(1\right)\) ) 

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