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\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)
c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)
\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)
\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)
\(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x\left(x^2+2-2x\right)-\left(x^2+2-2x\right)}\)
\(=\dfrac{x^2+2+2x}{x-1}\)
Bài 2:
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{10}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{1}{x+1}\)
c) Trong ngoặc giữa hai phân số là dấu gì vậy ?
a) \(\dfrac{x^2-4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x-2\right)}\)
\(=\dfrac{x+2}{2x}\)
b) \(\dfrac{2x-x^2}{x^2-4x+4}\)
\(=\dfrac{x\left(2-x\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(2-x\right)}{\left(2-x\right)^2}\)
\(=\dfrac{x}{2-x}\)
c) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\dfrac{x-y}{x+y}\)
d) \(\dfrac{5x^2+10x+5}{x+x^2}\)
\(\dfrac{5\left(x^2+2x+1\right)}{x\left(1+x\right)}\)
\(=\dfrac{5\left(x+1\right)^2}{x\left(x+1\right)}\)
\(=\dfrac{5\left(x+1\right)}{x}\)
e) \(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x+6\right)}\)
\(=\dfrac{3x\left(x+1\right)}{\left(x+1\right).2\left(x+3\right)}\)
\(=\dfrac{3x}{2\left(x+3\right)}\)
f) \(\dfrac{\left(2-3x\right)\left(x+1\right)}{x^2+2x+1}\)
\(=\dfrac{\left(2-3x\right)\left(x+1\right)}{\left(x+1\right)^2}\)
\(=\dfrac{2-3x}{x+1}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a.\(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\) : \(\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)^2}\)
= \(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\). \(\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)
b. \(\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\). \(\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}\)
= \(\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
c.Tương tự hai câu trên nka!!
d. (\(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2-x}{x+1}\)).(\(\dfrac{x}{x-1}\))
=( \(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2x-x^2}{x\left(x+1\right)}\)). ....
= \(\dfrac{\left(1-x\right)^2}{x\left(x+1\right)}\). ...
= \(\dfrac{x-1}{x+1}\)
\(\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\\ =\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)^2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\\ =\left(\dfrac{2x}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}\right)\\ =\dfrac{2x}{\left(x+2\right)^2}:\dfrac{-x}{x^2-4}\\ =\dfrac{2x}{\left(x+2\right)^2}.\dfrac{x^2-4}{-x}=-\dfrac{2x^2-8}{\left(x+2\right)^2}\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
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 \(x=-2\)
ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)
\(P=\dfrac{x+2}{x^2-4x+4}:\left(\dfrac{6-x^2}{x^2-2x}-\dfrac{1}{2-x}+\dfrac{x+2}{x}\right)\)
\(=\dfrac{x+2}{\left(x-2\right)^2}:\left(\dfrac{6-x^2}{x\left(x-2\right)}+\dfrac{1}{x-2}+\dfrac{x+2}{x}\right)\)
\(=\dfrac{x+2}{\left(x-2\right)^2}:\dfrac{6-x^2+x+\left(x+2\right)\left(x-2\right)}{x\left(x-2\right)}\)
\(=\dfrac{x+2}{\left(x-2\right)^2}\cdot\dfrac{x\left(x-2\right)}{6-x^2+x+x^2-4}\)
\(=\dfrac{x+2}{x-2}\cdot\dfrac{x}{x+2}=\dfrac{x}{x-2}\)