Question

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

Mn giúp mk vs ạ!!! Ai lm đúng mk sẽ tick cho

Thanks mn nhiều......

Answer 1

\(VT=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\) \(\Rightarrow VT\ge4\) (1)

Lại có \(3\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{8}{3\left(x+1\right)^2+2}\le\dfrac{8}{2}=4\) \(\Rightarrow VP\le4\) (2)

Từ (1), (2) \(\Rightarrow VT\ge VP\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\left|2x+3\right|+\left|2x-1\right|=4\\\dfrac{8}{3\left(x+1\right)^2+2}=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x+3\right)\left(1-2x\right)\ge0\\3\left(x+1\right)^2=0\end{matrix}\right.\) \(\Rightarrow x=-1\)

Vậy pt có nghiệm duy nhất \(x=-1\)

Answer 2

Mong mn giúp

Mk sắp thi r ạ!!!