Đặt \(t=x-7\) ta có:
\(\left(x-6\right)^4+\left(x-8\right)^4=16\Leftrightarrow\left(t+1\right)^4+\left(t-1\right)^2=16\)
\(\Leftrightarrow t^4+4t^3+6t^2+4t+1+\left(t^4-4t^3+6t^2-4t+1\right)=16\)
\(\Leftrightarrow2t^4+12t^2+2=16\Leftrightarrow t^4+6t^2-7=0\)
\(\Leftrightarrow\left(t^2-1\right)\left(t^2+7\right)=0\Leftrightarrow t^2-1=0\) ( do \(t^2+7>0\forall t\) )
\(\Leftrightarrow\left[{}\begin{matrix}t=1\Rightarrow x-7=1\Rightarrow x=8\\t=-1\Rightarrow x=6\end{matrix}\right.\)
\(\left(x-6\right)^4+\left(x-8^4\right)=16\)
=> \(\left(x-6\right)^4+\left(x-8\right)^4=2^4\) và \(-2^4\)
=> x - 6 + x - 8 = 2 và x - 6 + x - 8 = -2
=> 2x -14 = 2 2x - 14 = -2
2x = 16 2x = 12
x = 8 x= 6
thôi mk ra rồi để mk tự giải:
Đặt a= x-7, ta có:
\(\left(a+1\right)^4+\left(a-1\right)^4=16\)
\(\Leftrightarrow\)\(\left(a^4+4a^3+6a^2+4a+1\right)+\left(a^4-4a^3+6a^2-4a+1\right)=16\)
\(\Leftrightarrow\)\(2a^4+12a^2+2=16\)
\(\Leftrightarrow\)\(a^4+6a^2+1=8\)
\(\Leftrightarrow\)\(a^4+6a^2-7=0\)
\(\Leftrightarrow\)\(\left(a^4-1\right)+\left(6a^2-6\right)=0\)
\(\Leftrightarrow\)\(\left(a^2-1\right)\left(a^2+1\right)+6\left(a^2-1\right)=0\)
\(\Leftrightarrow\)\(\left(a^2-1\right)\left(a^2+7\right)=0\)
\(\Leftrightarrow\)\(\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)
Mà \(a^2+7>0\)
\(\Rightarrow\)\(\left(a-1\right)\left(a+1\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=-1\\a=1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-7=-1\\x-7=1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=6\\x=8\end{matrix}\right.\)
