Question

[Thầy Cao Đô]

Câu 3. (1,5 điểm): Tìm $x$, biết:

a) $(x+2)^2-x(x-1)=10$;

b) $x^3-6 x^2+9 x=0$.

Answer 1

[GV Nguyễn Trần Thành Đạt]

\(a,\\ \left(x+2\right)^2-x.\left(x-1\right)=10\\ \Leftrightarrow x^2+4x+4-x^2+x=10\\ \Leftrightarrow\left(x^2-x^2\right)+4x+x=10-4\\ \Leftrightarrow5x=6\\ \Leftrightarrow x=\dfrac{6}{5}\\ b,\\ x^3-6x^2+9x=0\\ \Leftrightarrow x.\left(x^2-6x+9\right)=0\\ \Leftrightarrow x.\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Answer 2

[Đặng Thảo Nguyên]