Question

Tìm giá trị x thỏa mãn: \(|x+3|+|x-1|=\frac{8}{3.\left(x+2\right)^2+2}\)

Answer 1

Ta có:\(\left(x+2\right)^2\ge0\)

\(\Rightarrow3\left(x+2\right)^2\ge0\)

\(\Rightarrow3\left(x+2\right)^2+2\ge2\)

\(\Rightarrow\frac{8}{3\left(x+2\right)^2+2}\le4\left(1\right)\)

Ta lại có:

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

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

Dấu "=" xảy ra khi và chỉ khi:\(\left(x+3\right)\left(1-x\right)\ge0\)

\(\Leftrightarrow1\le x\le3\left(3\right)\)

Từ (1),(2) ta có:\(\frac{8}{3\left(x+2\right)^2+2}=4\)

\(\Leftrightarrow8=12\left(x+2\right)^2+8\)

\(\Leftrightarrow\left(x+2\right)^2=0\)

\(\Leftrightarrow x=-2\)

Thay x vào (3) ta thấy thỏa mãn 

Vậy \(x=-2\)