1, Phân tích đa thức thành nhân tử: x3 - x2 +x +3
2, Cho
/a/<1;/a-c/<1999;/b-c/<1999. Chứng minh rằng /ab - c/<3998
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)\)
\(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\)
a, \(x^2\) + 4\(x\) - y2 + 4
= (\(x^2\) + 4\(x\) + 4) - y2
= (\(x\) + 2)2 - y2
= (\(x\) + 2 - y)(\(x\) + 2 + y)
b, 2\(x^2\) - 18
= 2.(\(x^2\) -9)
= 2.(\(x\) -3).(\(x\) + 3)
\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2\)
\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)\)
a) \(A=x^2-6x+9-9y^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3-3y\right)\left(x-3+3y\right)\)
b) \(B=x^3-3x^2+3x-1+2\left(x^2-1\right)\)
\(=\left(x-1\right)^3+\left(2x+2\right)\left(x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)^2+2x+2\right]\)
\(=\left(x-1\right).\left(x^2+3\right)\)
a, \(A=\left(x-3\right)^2-9y^2=\left(x-3-3y\right)\left(x-3+3y\right)\)
b, \(B=\left(x-1\right)^3+2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left[\left(x-1\right)^2+2\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2-2x+1+2x+2\right)=\left(x-1\right)\left(x^2+3\right)\)
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)