a) \(\left(x-4\right)^4+\left(x-2\right)^4=82\)
Đặt \(x-3=a\), ta có:
\(\left(a-1\right)^4+\left(a+1\right)^4=82\)
\(\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=82\)
\(\Rightarrow\left(a^2-2a+1\right)^2+\left(a+2a+1\right)^2=82\)
\(\Rightarrow\left(a^2+1\right)^2-4a\left(a^2+1\right)+4a^2+\left(a^2+1\right)^2+4a\left(a^2+1\right)+4a^2=82\)
\(\Rightarrow2\left[\left(a^2+1\right)^2+4a^2\right]=82\)
\(\Rightarrow\left(a^2+1\right)^2+4a^2=41\)
\(\Rightarrow a^4+2a^2+1+4a^2=41\)
\(\Rightarrow a^4+6a^2+1=41\)
\(\Rightarrow a^4+6a^2-40=0\)
\(\Rightarrow a^4-4a^2+10a^2-40=0\)
\(\Rightarrow a^2\left(a^2-4\right)+10\left(a^2-4\right)=0\)
\(\Rightarrow\left(a^2-4\right)\left(a^2+10\right)=0\)
\(\Rightarrow\left(a-2\right)\left(a+2\right)\left(a^2+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-2=0\\a+2=0\\a^2+10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=-2\\a^2=-10\end{matrix}\right.\)
Vì a2 = -10 ( Không tồn tại )
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b)Sửa đề
\(6x^4+5x^3-38x^2+5x+6=0\)
Xét x = 0 không phải là nghiệm của phương trình
Xét \(x\ne0\), chia cả hai vế cho x2, ta được
\(6x^2+5x-38+\dfrac{5}{x}+\dfrac{6}{x^2}=0\)
\(\Rightarrow6\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-38=0\)
Đặt \(x+\dfrac{1}{x}=a\left(1\right)\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=a^2\)
\(\Rightarrow x^2+\dfrac{1}{x^2}+2=a^2\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=a^2-2\left(2\right)\)
Thay (1) và (2) vào phương trình
\(6\left(a^2-2\right)+5a-38=0\)
\(\Rightarrow6a^2-12+5a-38=0\)
\(\Rightarrow6a^2+5a-50=0\)
\(\Rightarrow6a^2-15a+20a-50=0\)
\(\Rightarrow3a\left(2a-5\right)+10\left(2a-5\right)=0\)
\(\Rightarrow\left(2a-5\right)\left(3a+10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2a-5=0\\3a+10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2a=5\\3a=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=-\dfrac{10}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5}{2}\\x+\dfrac{1}{x}=-\dfrac{10}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=-\dfrac{1}{3}\\x=-3\end{matrix}\right.\)
