Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
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 \(x=-2\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}=\dfrac{15x.2y^2}{7y^3.x^2}=\dfrac{30}{7xy}\)
b) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)=\dfrac{-4y^2.3x^2}{11x^4.8y}=\dfrac{-3y}{22x^2}\)
c) \(\dfrac{x^3-8}{5x+20}.\dfrac{x^2+4x}{x^2+2x+4}\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}.\dfrac{x\left(x+4\right)}{x^2+2x+4}\\ =\dfrac{x^2-2x}{5}\)
a.
\(\dfrac{x+3}{x-2}+\dfrac{4+x}{2-x}\\ =\dfrac{x+3}{x-2}-\dfrac{4+x}{x-2}\\ =\dfrac{x+3-4-x}{x-2}\\ =-\dfrac{1}{x-2}\)
b. \(\dfrac{x+1}{2x+6}+\dfrac{2x+3}{x^2+3x}\)
\(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x}{2x\left(x+3\right)}+\dfrac{4x+6}{2x\left(x+3\right)}=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x^2+3x+2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x\left(x+3\right)+2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{x+2}{2x}\)
c. \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
d. \(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)
\(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}:\dfrac{-x\left(x+3\right)}{3x-1}\)
\(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}.\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}\)
\(=-\dfrac{2}{x^2}\)
1) \(\dfrac{x}{x+1}-\dfrac{2x}{x-1}+\dfrac{x+3}{x^2-1}\)
\(=\dfrac{x}{x+1}-\dfrac{2x}{x-1}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\) MTC: \(\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x-1\right)-2x\left(x+1\right)+\left(x+3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-x-2x^2-2x+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2-2x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+x-3x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x^2-x\right)-\left(3x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x\left(x-1\right)-3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)\left(-x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-3}{x+1}\)
2) \(\dfrac{5}{x+1}-\dfrac{10}{x-x^2-1}-\dfrac{15}{x^3+1}\)
\(=\dfrac{5}{x+1}-\dfrac{10}{-\left(x^2-x+1\right)}-\dfrac{15}{x^3+1}\)
\(=\dfrac{5}{x+1}+\dfrac{10}{\left(x^2-x+1\right)}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{10\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x}{x^2-x+1}\)
3) \(\dfrac{2}{2x+1}-\dfrac{1}{2x-1}-\dfrac{2}{1-4x^2}\)
\(=\dfrac{2}{2x+1}-\dfrac{1}{2x-1}+\dfrac{2}{4x^2-1}\)
\(=\dfrac{2}{2x+1}-\dfrac{1}{2x-1}+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\) MTC: \(\left(2x-1\right)\left(2x+1\right)\)
\(=\dfrac{2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2x+1}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{2\left(2x-1\right)-\left(2x+1\right)+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{4x-2-2x-1+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{2x-1}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{1}{2x+1}\)
4) \(\dfrac{3x^2+5x+14}{x^3+1}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\)
\(=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(3x^2+5x+14\right)+\left(x-1\right)\left(x+1\right)-4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9x+9}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9}{x^2-x+1}\)
a.\(\frac{5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(1-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\) = \(\frac{5x+1-1+3x-2x^2+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\) =\(\frac{10x+2}{x^3-1}\)
b.\(\frac{5}{x+1}+\frac{10}{x^2-x+1}-\frac{15}{x^3+1}\)( đến đây dễ r đúng ko)
a, \(\dfrac{4}{x^2-4}-\dfrac{2x}{x^2-4}=\dfrac{4-2x}{x^2-4}=\dfrac{-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\dfrac{2}{x+2}\)
\(b,\dfrac{3x+5}{x^2-5x}+\dfrac{x-25}{5x-25}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)
\(=\dfrac{5\left(3x+5\right)}{5x\left(x-5\right)}+\dfrac{\left(x-25\right)x}{5x\left(x-5\right)}\)
\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}\)
\(=\dfrac{x^2-10x+25}{5x\left(x-5\right)}\)
\(=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)
\(c,\left(\dfrac{2}{x-1}-\dfrac{2}{x+1}\right).\dfrac{x^2+2x+1}{4}\)
\(=\left(\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{\left(x+1\right)^2}{4}\)
\(=\dfrac{2x+2-2x+2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x+1\right)^2}{4}\)
\(=\dfrac{4}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x+1\right)^2}{4}\)
\(=\dfrac{x+1}{x-1}\)
a) \(\dfrac{x^2+2x}{x+2}=\dfrac{x\left(x+2\right)}{x+2}=x\)
b) \(\dfrac{5x+4-3\left(x-2\right)}{3\left(x+5\right)}=\dfrac{5x+4-3x+6}{3\left(x+5\right)}=\dfrac{2x+10}{3\left(x+5\right)}=\dfrac{2\left(x+5\right)}{3\left(x+5\right)}=\dfrac{2}{3}\)