1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)

\(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\)

\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\)

\(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)

b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\)

\(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\)

\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

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 \(x=-2\)

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) \(\dfrac{3x-2}{2xy}+\dfrac{7x+2}{2xy}\)

\(=\dfrac{\left(3x-2\right)+\left(7x+2\right)}{2xy}\)

\(=\dfrac{3x-2+7x+2}{2xy}\)

\(=\dfrac{10x}{2xy}\)

\(=\dfrac{5}{y}\)

b) \(\dfrac{5x+y^2}{x^2y}+\dfrac{x^2-5y}{xy^2}\) MTC: \(x^2y^2\)

\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}+\dfrac{x\left(x^2-5y\right)}{x^2y^2}\)

\(=\dfrac{y\left(5x+y^2\right)+x\left(x^2-5y\right)}{x^2y^2}\)

\(=\dfrac{5xy+y^3+x^3-5xy}{x^2y^2}\)

\(=\dfrac{y^3+x^3}{x^2y^2}\)

c) \(\dfrac{3x-2}{2xy}-\dfrac{7x-y}{2xy}\)

\(=\dfrac{\left(3x-2\right)-\left(7x-y\right)}{2xy}\)

\(=\dfrac{3x-2-7x+y}{2xy}\)

\(=\dfrac{-2-4x+y}{2xy}\)

d) \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\) MTC: \(x^2y^2\)

\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)

\(=\dfrac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)

\(=\dfrac{y^3+x^3}{x^2y^2}\)

e) \(\dfrac{16xy}{3x-1}.\dfrac{3-9x}{12xy^3}\)

\(=\dfrac{16xy\left(3-9x\right)}{12xy^3\left(3x-1\right)}\)

\(=\dfrac{4\left(3-9x\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-4\left(9x-3\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-4.3\left(3x-1\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-12}{3y^2}\)

\(=\dfrac{-4}{y^2}\)

f) \(\dfrac{8xy}{3x-1}:\dfrac{12xy^3}{5-15x}\)

\(=\dfrac{8xy}{3x-1}.\dfrac{5-15x}{12xy^3}\)

\(=\dfrac{8xy\left(5-15x\right)}{12xy^3\left(3x-1\right)}\)

\(=\dfrac{2\left(5-15x\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-2\left(15x-5\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-2.5\left(3x-1\right)}{3y^2\left(3x-1\right)}\)

\(=\dfrac{-10}{3y^2}\)

a: \(\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)

3/x^2-9=6/2(x+3)(x-3)

b: \(\dfrac{2x}{x^2-8x+16}=\dfrac{2x}{\left(x-4\right)^2}=\dfrac{6x^2}{3x\left(x-4\right)^2}\)

\(\dfrac{x}{3x^2-12x}=\dfrac{x}{3x\left(x-4\right)}=\dfrac{x\left(x-4\right)}{3x\left(x-4\right)^2}\)

c: \(\dfrac{x+y}{x}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{x\left(x-y\right)}\)

x/x-y=x^2/x(x-y)

e: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(x+2\right)\left(2-x\right)}\)

\(\dfrac{8}{2x-x^2}=\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(2+x\right)}\)

Giúp mình nhé mọi người !

\(1.\)

\(a.\)

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=x-1\)

\(b.\)

\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)

\(=\dfrac{2y}{\left(x-y\right)}\)

Tương tự các câu còn lại