Đề có ghi gì đâu mà bất định với có định :VV

"Dùng phương pháp hệ số bất định" để làm gì?

c) \(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)

\(=\left(x+1\right)^4+x^4+x^2+1+2x^3+2x^2+2x\)

\(=\left(x+1\right)^4+x^4+3x^2+1+2x^3+2x\)

a) \(x^4-7x^3+14x^2-7x+1\)(1)

Giả sử x khác 0, khi đó :

\(\left(1\right)\Leftrightarrow x^2\left(x^2-7x+14-\dfrac{7}{x}+\dfrac{1}{x^2}\right)\)

\(\Leftrightarrow x^2\left[\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+14\right]\)

\(\Leftrightarrow x^2\left[\left(x^2+2\cdot x\cdot\dfrac{1}{x}+\dfrac{2}{x^2}\right)-2-7\left(x+\dfrac{1}{x}\right)+14\right]\)

\(\Leftrightarrow x^2\left[\left(x+\dfrac{1}{x}\right)^2-7\left(x+\dfrac{1}{x}\right)+12\right]\)

Đặt \(x+\dfrac{1}{x}=a\)

pt \(\Leftrightarrow x^2\left(a^2-7a+12\right)\)

\(\Leftrightarrow x^2\left(a^2-3a-4a+12\right)\)

\(\Leftrightarrow x^2\left[a\left(a-3\right)-4\left(a-3\right)\right]\)

\(\Leftrightarrow x^2\left(a-3\right)\left(a-4\right)\)

\(\Leftrightarrow x^2\left(x+\dfrac{1}{x}-3\right)\left(x+\dfrac{1}{x}-4\right)\)

a: \(=2x^3+x^2-2x^2-x+6x+3\)

\(=\left(2x+1\right)\left(x^2-x+3\right)\)

b: \(=x^3+x^2+4x^2+4x+4x+4\)

\(=\left(x+1\right)\left(x^2+4x+4\right)\)

\(=\left(x+1\right)\left(x+2\right)^2\)

c: \(=\left(x^2+7x+12\right)\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)