a, \(x^2-8y^2+6x+9\)
\(=\left(x+3\right)^2-8y^2\)
\(=\left(x+3-\sqrt{8}y\right)\left(x+3+\sqrt{8}y\right)\)
b, \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left[\left(4x+1\right)\left(3x+2\right)\right]\left[\left(12x-1\right)\left(x+1\right)\right]-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)(1)
Đặt \(12x^2+11x+2=a\Rightarrow12x^2+11x-1=a-3\)
\(\Rightarrow\left(1\right)=a\left(a-3\right)-4=a^2-3a-4\)
\(=a^2+a-4a-4=a\left(a+1\right)-4\left(a+1\right)\)
\(=\left(a+1\right)\left(a-4\right)\)(*)
Vì \(a=12x^2+11x+2\) nên:
\(\left(\text{*}\right)=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
Chúc bạn học tốt!!!
a) Sửa đề:\(x^2-9y^2+6x+9\) (đúng chưa cậu?)
\(=x^2+6x+9-9y^2\)
\(=\left(x+3\right)^2-\left(3y\right)^2\)
\(=\left(x+3-3y\right)\left(x+3+3y\right)\)
b) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(4x+1\right)\left(3x+2\right)\left(12x-1\right)\left(x+1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x+2=t\)
\(t\left(t-3\right)-4=t^2-3t-4\)
\(=t^2-4t+t-4\)
\(=t\left(t-4\right)+\left(t-4\right)\)
\(=\left(t-4\right)\left(t+1\right)\)
Thay t, đa thức có thể được phân tích thành
\(\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
