Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=xy\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(xy-7\right)\)
a) \(24x^2-4xy\)
\(=4x\left(6x-y\right)\)
b) \(5x^3-10x^2+5x-20xy^2\)
\(=5x\left(x^2-10x+5-20y^2\right)\)
y^2 - ( x^2 - 2x + 1 )
⇔ y^2 - ( x - 1 )^2
⇔ ( y - x - 1 ) ( y + x - 1 )
nha bạn
( x + y )2 - 2( x + y ) + 1 =
= ( x + y - 1 )2 ( áp dụng hằng đẳng thức thứ bình phương 1 hiệu nha )
Hok tốt!!!!!!!!!!!!!!!!
\(c,Sửa:\dfrac{1}{4x+2}=\dfrac{x\left(2x-1\right)}{2x\left(2x+1\right)\left(2x-1\right)}\\ \dfrac{20}{4x^3-x}=\dfrac{40}{2x\left(2x+1\right)\left(2x-1\right)}\\ \dfrac{7}{2x^2+x}=\dfrac{7\left(2x-1\right)}{x\left(2x+1\right)\left(2x-1\right)}\\ d,\dfrac{x}{x-y};\dfrac{x^2-y^2}{x^2-2xy+y^2}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^2}=\dfrac{x+y}{x-y};x+y=\dfrac{x^2-y^2}{x-y}\\ e,\dfrac{x}{x+1}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)};\dfrac{x^2}{1-x}=\dfrac{-x^2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)};\dfrac{1}{x^2-1}=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)
a: \(5x-20y=5\left(x-4y\right)\)
b: \(x^2+x^2y+x^2y^2=x^2\left(1+y+y^2\right)\)
c: \(x\left(x+y\right)-\left(5x+5y\right)=\left(x+y\right)\left(x-5\right)\)
d: \(5\left(x-y\right)+y\left(x-y\right)=\left(x-y\right)\left(y+5\right)\)
\(2x^2+5x-3=\left(2x^2-x\right)+\left(6x-3\right)\)\(=x\left(2x-1\right)+3\left(2x-1\right)=\left(x+3\right)\left(2x-1\right)\)
d: =(x+2y-6)(x+2y+6)
c)
=x(4x-7)-2(4x-7)
=(x-2)(4x-7)