Điều kiện: \(x\ne1\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x-2}{x^2+x+1}\)
Vì \(x\ne1\)nên:
\(\Leftrightarrow\frac{x-2}{x^2+x+1}=\frac{x-2}{x^2+x+1}\)Đúng với mọi \(x\ne1\).ĐPCM
cách lm dễ nhất là bn nhân chéo.....roi phân tích đa thức thành nhân tử là đc
điều kiện x # 1
\(VT=\frac{\left(x-1\right).\left(x-2\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{x-2}{x^2+x+1}=VP\)(đpcm)
Ta có: \(\left(x^2-3x+2\right).\left(x^2+x+1\right)\)\(=x^2.\left(x^2+x+1\right)-3x.\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)
\(=x^4+x^3+x^2-3^3-3x^2-3x+2x^2+2x+2\)
\(=x^4-2x^3-x+2\) (1)
\(\left(x^3-1\right).\left(x-2\right)=x^3.\left(x-2\right)-\left(x-2\right)=x^4-2x^3-x+2\)(2)
Từ (1) và (2) suy ra: \(\left(x^2-3x+2\right).\left(x^2+x+1\right)\)=\(\left(x^3-1\right).\left(x-2\right)\)
=> \(\frac{x^2-3x+2}{x^3-1}=\frac{x-2}{x^2+x+1}\)(đpcm)
Ta có:
x2 - 3x + 2 / x3 - 1
= x2 - 2x - x + 2 / x3 - x2 + x2 - 1
= x.(x - 2) - (x - 2) / x2.(x - 1) + (x2 - 12)
= (x - 2).(x - 1) / x2.(x - 1) + (x - 1).(x + 1)
= (x - 2).(x - 1) / (x - 1).(x2 + x + 1)
= x - 2 / x2 + x + 1
Chứng tỏ x2 - 3x + 2 / x3 - 1 = x - 2 / x2 + x + 1
Ta có:
x2 - 3x + 2 / x3 - 1
= x2 - 2x - x + 2 / x3 - x2 + x2 - 1
= x.(x - 2) - (x - 2) / x2.(x - 1) + (x2 - 12)
= (x - 2).(x - 1) / x2.(x - 1) + (x - 1).(x + 1)
= (x - 2).(x - 1) / (x - 1).(x2 + x + 1)
= x - 2 / x2 + x + 1
Chứng tỏ x2 - 3x + 2 / x3 - 1 = x - 2 / x2 + x + 1
