Đề 1

Câu 1:

\(\text{a) }8x-4y=4\left(2x-y\right)\)

\(\text{b) }5y\left(x-1\right)-3x\left(x-1\right)=\left(5y-3x\right)\left(x-1\right)\)

\(\text{c) }-6x-6y+x\left(x+y\right)\\ =-\left(6x+6y\right)-x\left(x+y\right)\\ =-6\left(x+y\right)+x\left(x+y\right)\\ =\left(x-6\right)\left(x+y\right)\)

\(\text{d) }x^2-49=\left(x-7\right)\left(x+7\right)\)

Câu 2:

\(\text{a) }5x^2-x=0\\ \Leftrightarrow5x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ \text{Vậy }x=0\text{ hoặc }x=1\)

\(\text{b) }x^2-25=0\\ \Leftrightarrow x^2=25\\ \Leftrightarrow x=\pm5\\ \text{Vậy }x=\pm5\)

\(\text{c) }x^4+2x^3+6x-9=0\\ \Leftrightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\\ \Leftrightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^3+3x^2\right)+\left(3x+9\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x^2=-3\left(Vô\text{ }lí\right)\\x=-3\left(TM\right)\end{matrix}\right.\\ \text{Vậy }x=1\text{ }\text{hoặc }x=-3\)