Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2\)
\(=3xy\left(x+y\right)\)
\(a,\left(2x+3y\right)^2-4\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x+3y-4\right)\)
\(b,\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x^3+y^3\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-\left(x^2-xy+y^2\right)\right]\)
\(=\left(x+y\right).3x\)
\(c,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
b, \(2x+4x^2=5-5+4-2^2\)
\(\Leftrightarrow2x\left(1+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\1+2x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
Vậy...
\(2x+4x^2=5-5+4-2^2\)
\(\Leftrightarrow2x+4x^2=0\)
\(\Leftrightarrow4x^2+2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
Vậy:...
1) (x-2)3
2) (x-1+5x)(x-1-5x)=(6x-1)(-4x-1)
3) (6x-2x-1)(6x+2x+1)=(4x-1)(8x+1)
4) (x+3-2x+1)(x+3+2x-1) = (4-x)(3x+2)
1. \(x^3-6x^2+12x-8=\left(x-2\right)^3\)
2. \(\left(x-1\right)^2-25x^2=\left(x-1-5x\right)\left(x-1+5x\right)\)
= \(\left(-4x-1\right)\left(6x-1\right)\)
3. \(36x^2-\left(2x+1\right)^2=\left(6x-2x-1\right)\left(6x+2x+1\right)\)
= \(\left(4x-1\right)\left(8x+1\right)\)
4. \(\left(x+3\right)^2-\left(2x-1\right)^2=\left(x+3-2x+1\right)\left(x+3+2x-1\right)\)
= \(\left(4-x\right)\left(3x+2\right)\)
Chúc bạn làm bài tốt
x3-2x2-25xy2+x
=x(x2-2x+1-25y2)
=x[(x-1)2-25y2]
=x(x-1-5y)(x-1+5y)