a/ \(\left(x+1\right)\left(x+5\right)\left(x^2+6x+19\right)=0\)
b/ \(\left(x+1\right)\left(x^2-2x+2\right)\left(x^2+x+1\right)=0\)
e/ \(\left(x+3\right)\left(x+5\right)\left(x^2+9x+19\right)=0\)
a) \(\left(x+2\right)^4+\left(x+4\right)^4=82\)
x+3=t
<=>\(\left(t-1\right)^4+\left(t+1\right)^4=82\)
<=>\(\left[\left(t-1\right)^2-\left(t+1\right)^2\right]^2=82-2\left(t-1\right)^2\left(t+1\right)^2\)
<=>\(\left[\left\{\left(t-1\right)-\left(t+1\right)\right\}\left\{\left(t-1\right)+\left(t+1\right)\right\}\right]^2=82-2\left(t^2-1\right)^2\)
<=>\(16t^2=82-2\left(t^2-1\right)^2\)
<=>\(\left(t^2-1\right)^2+8t^2-41=0\)
<=>\(\left(t^2-1\right)^2+8\left(t^2-1\right)-33=0\)
\(\Delta_{\left(t^2-1\right)}=16+33=49\)
\(\left[{}\begin{matrix}t^2-1=-4-7\left(l\right)\\t^2-1=-4+7\Leftrightarrow t^2=4\Rightarrow\left[{}\begin{matrix}t_1=2\\t_2-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=-5\\x_2=-1\end{matrix}\right.\)
d) Đặt \(y=x+1\)
\(\Rightarrow pt\Leftrightarrow\left(y+1\right)^3-3\left(y+1\right)^2+9\left(y+1\right)-9=0\\ \Leftrightarrow y^3+6y-2=0\left(2\right)\)
Đặt \(y=u+v\), tìm được u và v sao cho \(\left\{{}\begin{matrix}u^3+v^3=2\\u^3v^3=-8\end{matrix}\right.\)
Khi đó \(u^3\) và \(v^3\) là 2 nghiệm của pt: \(X^2-2X-8=0\left(3\right)\)
Giải (3) được \(X_1=4;X_2=-2\)
\(\Rightarrow x=u+v+1=\sqrt[3]{4}-\sqrt[3]{2}+1\)