Question

phân tích đa thức thành nhân tử

(x-2).(x-3).(x-4).(x-5)+1

x.(y+z)²+y.(z+X)²+z.(x+y)²-3xyz

Answer 1

a) \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)

\(=\left[\left(x-2\right)\left(x-5\right)\right]\left[\left(x-3\right)\left(x-4\right)\right]+1\)

\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\) 

Đặt: \(x^2-7x+11=t\)

\(\Rightarrow\hept{\begin{cases}x^2-7x+10=t-1\\x^2-7x+12=t+1\end{cases}}\)

\(\Rightarrow\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)

\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)

\(=\left(t-1\right)\left(t+1\right)+1\)

\(=t^2-1+1\)

\(=t^2\)

Vậy: \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)

\(=\left(x^2-7x+11\right)^2\)