Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) \(\left(x+\dfrac{1}{2}\right)^2-2x^2\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-2x^2\)
\(=x^2+x+\dfrac{1}{4}-2x^2\)
\(=-x^2+x+\dfrac{1}{4}\)
b) \(\left(x-2y\right)^2-4y^2\)
\(=x^2-2\cdot x\cdot2y+\left(2y\right)^2-4y^2\)
\(=x^2-4xy+4y^2-4y^2\)
\(=x^2-4xy\)
c) \(\left(x+\dfrac{1}{2}y\right)^3\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}y+3\cdot x+\left(\dfrac{1}{2}y\right)^2+\left(\dfrac{1}{2}y\right)^3\)
\(=x^3+\dfrac{3}{2}x^2y+\dfrac{3}{4}xy^2+\dfrac{1}{8}y^3\)
d) \(\left(2x^2-3y\right)^3\)
\(=\left(2x^2\right)^3-3\cdot\left(2x^2\right)^2\cdot3y+3\cdot2x^2\cdot\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^6-36x^4y+54x^2y^2-27y^3\)
e) \(\left(x^2+y\right)^2-\left(x+y\right)^2\)
\(=\left[\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\right]-\left(x^2+2\cdot x\cdot y+y^2\right)\)
\(=\left(x^4+2x^2y+y^2\right)-\left(x^2+2xy+y^2\right)\)
\(=x^4+2x^2y+y^2-x^2-2xy-y^2\)
\(=x^4+2x^2y-x^2-2xy\)
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
a/ \(\dfrac{4x+2}{3x^2-x}:\dfrac{x^2+3x}{1-3x}=-\dfrac{4x+2}{x\left(1-3x\right)}\cdot\dfrac{1-3x}{x^2+3x}=-\dfrac{4x^2+2}{x\left(x^2+3x\right)}\)
b/ \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2-12xy+9y^2}{1-x^2}=-\dfrac{2\left(2x+3y\right)}{1-x}\cdot\dfrac{\left(1-x\right)\left(1+x\right)}{\left(2x+3y\right)^2}=\dfrac{-2\left(x+1\right)}{2x+3y}=\dfrac{-2x-2}{2x+3y}\)
c/ \(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}=\dfrac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}\cdot\dfrac{2x+y}{x\left(x^2+xy+y^2\right)}=\dfrac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y}\cdot\dfrac{1}{x\left(x^2+xy+y^2\right)}=\dfrac{x-y}{y}\)
\(a,x^3-6x^2y+12xy^2-8x^3=\left(x-2y\right)^3\)
\(b,x^2+2x+1-4x^2=\left(x+1\right)^2-\left(2x\right)^2=\left(x+1+2x\right)\left(x+1-2x\right)=\left(3x+1\right)\left(1-x\right)\)
câu a 8y^2 hay 3