Question

[Phạm Đức Thịnh]

Giải pt : \(2x^4+\left(1-2x\right)^4=\frac{1}{27}\)

Ai nhanh mk tick :D 

Answer 1

[9 Quả Chuổi 9]

\(2x^4+\left(1-2x\right)^4=\frac{1}{27}\)

\(\Rightarrow2x^4+\left(1-2x\right)^4=\frac{1}{3}\cdot3\cdot\left[2x^4+\left(1-2x\right)^4\right]\)

\(\Rightarrow\frac{1}{3}\left(1^2+1^2+1^2\right)\left[x^4+x^4+\left(1-2x\right)^4\right]\)\(\ge\)\(\frac{1}{3}\left[x^2+x^2+\left(1-2x\right)^2\right]^2\)

\(\Rightarrow\frac{1}{3}\cdot\frac{1}{9}\left\{3.\left[x^2+x^2+\left(1-2x\right)^2\right]\right\}^2\)

\(\Rightarrow\frac{1}{27}\left\{\left(1^2+1^2+1^2\right)\left[x^2+x^2+\left(1-2x\right)^2\right]\right\}^2\)\(\ge\)\(\frac{1}{27}\left[x+x+\left(1-2x\right)\right]^4=\frac{1}{27}\)

Vậy phương trình có nghiệm khi dấu đẳng thức xảy ra 

\(\Leftrightarrow\hept{\begin{cases}x^2=x^2=\left(1-2x\right)^2\\x=x=1-2z\end{cases}}\Leftrightarrow x=\frac{1}{3}\)

KL : ................................................................... 

P/s : chưa chắc