Question

giúp câu 4 câu 6

Answer 1

6: \(x^4-2x^3+x=2\)

=>\(x^4-2x^3+x-2=0\)

=>\(x^3\left(x-2\right)+\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x^3+1\right)=0\)

=>\(\left[\begin{array}{l}x-2=0\\ x^3+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-1\end{array}\right.\)

4: \(\left(x^2-6x\right)^2-2\left(x-3\right)^2+2=0\)

=>\(\left(x^2-6x+9-9\right)^2-2\cdot\left(x-3\right)^2+2=0\)

=>\(\left\lbrack\left(x-3\right)^2-9\right\rbrack^2-2\left(x-3\right)^2+2=0\)

=>\(\left(x-3\right)^4-20\left(x-3\right)^2+83=0\)

=>\(\left\lbrack\left(x-3\right)^2-10\right\rbrack^2=17\)

=>\(\left(x-3\right)^2-10=\sqrt{17}\) hoặc \(\left(x-3\right)^2-10=-\sqrt{17}\)

=>\(\left[\begin{array}{l}\left(x-3\right)^2=10+\sqrt{17}\\ \left(x-3\right)^2=10-\sqrt{17}\end{array}\right.\)

TH1: \(\left(x-3\right)^2=10+\sqrt{17}\)

=>\(\left[\begin{array}{l}x-3=\sqrt{10+\sqrt{17}}\\ x-3=-\sqrt{10+\sqrt{17}}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3+\sqrt{10+\sqrt{17}}\\ x=3-\sqrt{10+\sqrt{17}}\end{array}\right.\)

TH2: : \(\left(x-3\right)^2=10-\sqrt{17}\)

=>\(\left[\begin{array}{l}x-3=\sqrt{10-\sqrt{17}}\\ x-3=-\sqrt{10-\sqrt{17}}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3+\sqrt{10-\sqrt{17}}\\ x=3-\sqrt{10-\sqrt{17}}\end{array}\right.\)