phân tích đa thức thành nhân tử : 3x - y( với x,y > 0)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(3x\cdot\left(x-y\right)^2-6\cdot\left(y-x\right)\)
\(=3x\left(x-y\right)^2+6\left(x-y\right)\)
\(=\left(x-y\right)\left[3x\left(x-y\right)+6\right]\)
\(=\left(x-y\right)\left(3x^2-3xy+6\right)\)
\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(x+y+3\right)\)
a) \(=3x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(3x-1\right)\)
c) \(2x\left(y-x\right)+3y\left(x-y\right)=\left(2x-3y\right)\left(y-x\right)\)
d) \(=3\left(x^2+2x+1-y^2\right)=3\left[\left(x+1\right)^2-y^2\right]=3\left(x-y+1\right)\left(x+y+1\right)\)
\(x^3+y^3-3x-3y=\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-3\right)\)
\(x^3+y^3-3x-3y\)
<=> \( \left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)\)
<=>\(\left(x+y\right)\left(x^2+y^2-xy-3\right)\)
\(=\left(\sqrt{2x}\right)^2-\left(\sqrt{y}\right)^2\)
\(=\left(\sqrt{2x}-\sqrt{y}\right)\left(\sqrt{2x}+\sqrt{y}\right)\)
3\(x\) - y
= (\(\sqrt{3x}\))2 - (\(\sqrt{y}\))2
= (\(\sqrt{3x}\) - \(\sqrt{y}\)).(\(\sqrt{3x}\) + \(\sqrt{y}\))