bài 1
a) \(7x\left(5x-1\right)+5x-1=\left(5x-1\right)\left(7x+1\right)\)
b) \(4xy-4x^2-y^2+25=25-\left(4x^2-4xy+y^2\right)\)
\(=5^2-\left(2x-y\right)=\left(5-2x+y\right)\left(5+2x-y\right)\)
c) \(2x^2-2y+xy-4x=\left(2x^2+xy\right)-\left(2y+4x\right)\)
\(=x^2\left(2x+y\right)-2\left(2x+y\right)=\left(2x+y\right)\left(x^2-2\right)\)
d) \(3x^2-7x+2=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
bài 2
a) * Rút gọn:
\(Q=3\left(2x-1\right)^2+2\left(2x+3\right)\left(x-1\right)-\left(x-3\right)\left(x+3\right)\)
\(Q=\left[3\left(4x^2-4x+1\right)\right]+\left[2\left(2x^2-2x+3x-3\right)\right]-\left(x^2-9\right)\)
\(Q=\left(12x^2-12x+3\right)+\left(4x^2-4x+6x-6\right)-\left(x^2-9\right)\)
\(Q=12x^2-12x+3+4x^2-4x+6x-6-x^2+9\)
\(Q=15x^2-10x+6=5x\left(3x-2\right)+6\)
Thế x = 2 vào biểu thức Q ta được:
\(Q=5\cdot2\left(3\cdot2-2\right)+6=46\)
b) \(Q=5x\left(3x-2\right)+6=6\)
\(\Leftrightarrow5x\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
