câu a bạn sai đề nha
b)
\(\left(x^2+x+1\right)^2=3\left(x^4+x^2+1\right)\)
\(x^4+x^2+1+2x^3+2x^2+2x=3x^4+3x^2+3\)
\(2\left(x^3+x^2+x\right)=2\left(x^4+x^2+1\right)\)
\(x^4-x^3+1-x=0\)
\(x^3\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^3-1\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\x^3-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)
a) (x2 - 5x)2 + 10(x2 - 5x) + 24 = 0
\(\Leftrightarrow\) (x2 - 5x)2 + 10(x2 - 5x) + 25 - 1 = 0
\(\Leftrightarrow\) [(x2 - 5x) + 5]2 - 1 = 0
\(\Leftrightarrow\) (x2 - 5x + 5 - 1)(x2 - 5x + 5 + 1) = 0
\(\Leftrightarrow\) (x2 - 5x + 4)(x2 - 5x + 6) = 0
\(\Leftrightarrow\) (x2 - 4x - x + 4)(x2 - 2x - 3x + 6) = 0
\(\Leftrightarrow\) [x(x - 4) - (x - 4)][x(x - 2) - 3(x - 2)] = 0
\(\Leftrightarrow\) (x - 4)(x - 1)(x - 2)(x - 3) = 0
\(\Leftrightarrow\) x - 4 = 0 hoặc x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0
* x - 4 = 0
\(\Leftrightarrow\) x = 4
* x - 1 = 0
\(\Leftrightarrow\) x = 1
* x - 2 = 0
\(\Leftrightarrow\) x = 2
* x - 3 = 0
\(\Leftrightarrow\) x = 3
Vậy S = \(\left\{1;2;3;4\right\}\)
