Rút gọn phân thức:
\(A=\dfrac{2012.2013+2014}{2010-2012.2015}\)
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\(A=\dfrac{2012\left(2012+1\right)+2012+2}{2012-2-2012\cdot2015}\)
\(=\dfrac{2012^2+2012\cdot2+2}{2012-2-2012\left(2012+3\right)}\)
\(=\dfrac{2012^2+4026}{-2012^2-4026}=-1\)
a/ \(\frac{2.2012.2014-2}{2011+2012.\left(2014-1\right)}=\frac{2.\left(2012.2014-1\right)}{2011+2012.2014-2012}\)
\(=\frac{2.\left(2012.2014-1\right)}{2012.2014-1}=2\)
b/ \(\frac{2012.2013+2014}{2010-2012.\left(2013+2\right)}=\frac{2012.2013+2014}{2010-2012.2013-4024}\)
\(=\frac{2012.2013+2014}{-\left(2012.2013+2014\right)}=-1\)
c/ \(\frac{6.11111.87564-3.11111}{2.11111\left(87564-4\right)+7.11111}=\frac{6.87564-3}{2.87564-8+7}\)
\(=\frac{3\left(2.87564-1\right)}{2.87564-1}=3\)
\(A=\left(1-\frac{1}{2011}\right)-\left(1-\frac{1}{2012}\right)+\left(1-\frac{1}{2013}\right)-\left(1-\frac{1}{2014}\right)\)
\(=1-\frac{1}{2011}-1+\frac{1}{2012}+1-\frac{1}{2013}-1+\frac{1}{2014}\)
\(=\left(1-1+1-1\right)-\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}+\frac{1}{2014}\right)\)
còn lại bó tay @@
\(A=\frac{2010}{2011}-\frac{2011}{2012}+\frac{2012}{2013}-\frac{2013}{2014}\)
và
\(B=\frac{1}{2010.2011}-\frac{1}{2012.2013}\)
3A=3/2.5+...+3/2018.2021
3A=1/2-1/5+1/5-...+1/2018-1/2021
3A=1/2-1/2021 sau tự tính A
a) ĐKXĐ:
\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
b) \(A=\dfrac{x^2-2x+1}{x^2-1}\)
\(A=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)
\(A=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\)
\(A=\dfrac{x-1}{x+1}\)
c) Thay x = 3 vào A ta có:
\(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
a) ĐKXĐ:
\(9x^2-y^2\ne0\Leftrightarrow\left(3x\right)^2-y^2\ne0\Leftrightarrow\left(3x-y\right)\left(3x+y\right)\ne0\)
\(\Leftrightarrow3x\ne\pm y\)
b) \(B=\dfrac{6x-2y}{9x^2-y^2}\)
\(B=\dfrac{2\cdot3x-2y}{\left(3x\right)^2-y^2}\)
\(B=\dfrac{2\left(3x-y\right)}{\left(3x+y\right)\left(3x-y\right)}\)
\(B=\dfrac{2}{3x+y}\)
Thay x = 1 và \(y=\dfrac{1}{2}\) và B ta có:
\(B=\dfrac{2}{3\cdot1+\dfrac{1}{2}}=\dfrac{2}{3+\dfrac{1}{2}}=\dfrac{2}{\dfrac{7}{2}}=\dfrac{4}{7}\)
\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x^2-4-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x-4}{x-2}\)
a. \(x^2-5x\ne0\)
=> ĐKXĐ: \(x\left(x-5\right)\ne0\) => \(\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)
b. \(\dfrac{x^2-10x+25}{x^2-5x}\)
= \(\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}\)
= \(\dfrac{x-5}{x}\)
\(A=\frac{2012.2013+2014}{2010-2012.2015}\)
\(=\frac{2012.2013+2014}{2010-2012.\left(2013+2\right)}\)
\(=\frac{2012.2013+2014}{2010-2012.2013-4024}\)
\(=\frac{2012.2013+2014}{-2012.2013-2014}=-1\)