Dài qá ak .-.
g) Đặt \(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt \(t=x^2+5x+5\)
⇒ \(A=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=\left(t+5\right)\left(t-5\right)\)
\(=x\left(x^2+5x+10\right)\left(x+5\right)\)
h) Đặt \(B=x^4-7x^3+14x^2-7x+1\)
\(=x^2\left(x^2-7x+14-\frac{7}{x}+\frac{1}{x^2}\right)\)
Đặt \(t=x+\frac{1}{x}\)
\(\Rightarrow t^2-2=x^2+\frac{1}{x}\)
Khi đó \(B=x^2\left(t^2-2-7t+14\right)\)
\(=x^2\left(t^2-7t+12\right)\)
\(=x^2\left(t^2-3t-4t+12\right)\)
\(=x^2\left(t-3\right)\left(t-4\right)\)
\(=x^2\left(x+\frac{1}{x}-3\right)\left(x+\frac{1}{x}-4\right)\)
\(=\left(x^2+1-3x\right)\left(x^2+1-4x\right)\)
Phân tích tiếp nhé <33
a) \(x^2+3x-18=\left(x^2-3x\right)+\left(6x-18\right)=x\left(x-3\right)+6\left(x-3\right)=\left(x-3\right)\left(x+6\right)\)
b) \(3x^2-16x+5=\left(3x^2-15x\right)+\left(-x+5\right)=3x\left(x-5\right)-\left(x-5\right)=\left(3x-1\right)\left(x-5\right)\)
c) \(x^3-5x^2+8x-4\)
\(=\left(x^3-4x^2+4x\right)+\left(-x^2+4x-4\right)\)
\(=x\left(x^2-4x+4\right)-\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)
d) \(x^3-7x+6\)
\(=\left(x^3+x^2-6x\right)+\left(-x^2-x+6\right)\)
\(=x\left(x^2+x-6\right)-\left(x^2+x-6\right)\)
\(=\left(x+1\right)\left(x^2+3x-2x+6\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x-2\right)\)
e) \(64x^4+1\)
\(=64x^4+16x^2+1-16x^2\)
\(=\left(4x^2+1\right)^2-16x^2\)
\(=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)
