Question

giải pt

`(x-3)^4 + (x-5)^4 =2`

Answer 1

Ta có:

\(\left(x-3\right)^4+\left(x-5\right)^4=2\)

\(\Leftrightarrow\left(x-4+1\right)^4+\left(x-4-1\right)^4=2\)

Đặt: \(y=x-4\) ta có: 

\(\Leftrightarrow\left(y+1\right)^4+\left(y-1\right)^4=2\)

\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=2\)

\(\Leftrightarrow2y^4+12y^2+2=2\)

\(\Leftrightarrow2y^4+12y^2=2-2\)

\(\Leftrightarrow2y^4+12y^2=0\)

\(\Leftrightarrow2y^2\left(y^2+6\right)=0\)

Mà: \(y^2+6\ge6>0\forall x\)

\(\Leftrightarrow2y^2=0\)

\(\Leftrightarrow y^2=0\)

\(\Leftrightarrow y=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)