Question

x²-5x+4

Answer 1

\(x^2-5x+4=0\Leftrightarrow x^2-4x-x+4=0\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right..Vậy:x\in\left\{1;4\right\}\)

tớ nghĩ đề là thế này

Answer 2

x^2 - 5x + 4= x^2 - 4x - x + 4 = x(x-4) - (x -4)= (x-4)(x-1)

Answer 3

\(x^2-5x+4\)

\(\Leftrightarrow x^2-x-4x+4\)

\(\Leftrightarrow\)\(x\left(x-1\right)-4\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\)