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1 tháng 12 2019
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
1 tháng 12 2019
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
4 tháng 7 2017
a. A=\(1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\frac{x+1+x+1-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right).\frac{x\left(x^2-x+1\right)}{x^2\left(x-2\right)}\)
\(=1+\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b.\(\left|x-\frac{3}{4}\right|=\frac{5}{4}\Rightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
Với \(x=2\Rightarrow A=\frac{2-1}{2+1}=\frac{1}{3}\)
Với \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
8 tháng 12 2019
a) Để P xác định \(\Leftrightarrow\hept{\begin{cases}2a-2\ne0\\2-2a^2\ne0\\a+2\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a^2\ne1\\a\ne-2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a\ne-1vâ\ne1\\a\ne-2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a\ne-1\\a\ne2\end{cases}}\)
b) \(P=\left(\frac{a+1}{2a-2}+\frac{1}{2-2a^2}\right).\frac{2a+2}{a+2}\)
\(=\left[\frac{a+1}{2\left(a-1\right)}+\frac{1}{2\left(1-a\right)\left(1+a\right)}\right].\frac{2\left(a+1\right)}{a+2}\)
\(=\left[\frac{\left(a+1\right)^2}{2\left(a-1\right)\left(a+1\right)}-\frac{1}{2\left(a-1\right)\left(1+a\right)}\right].\frac{2\left(a+1\right)}{a+2}\)
\(=\frac{\left(a+1\right)^2-1}{2\left(a-1\right)\left(a+1\right)}.\frac{2\left(a+1\right)}{a+2}\)
\(=\frac{a\left(a+2\right)}{\left(a-1\right)\left(a+2\right)}\)
\(=\frac{a}{a-1}\)
c) \(\left|a\right|=3\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)
+) Với a=3 thỏa mãn \(\hept{\begin{cases}a\ne1\\a\ne-1\\a\ne2\end{cases}}\)nên thay a=3 vào P ta được:
( làm nốt)
TH kia tương tự
LM
19 tháng 7 2017
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
DM
0
HN
24 tháng 12 2016
a) B xác định
\(\Leftrightarrow\begin{cases}2a^2+6a\ne0\\a^2-9\ne0\end{cases}\Leftrightarrow\begin{cases}2a\left(a+3\right)\ne0\\\left(a+3\right)\left(a-3\right)\ne0\end{cases}\Leftrightarrow\begin{cases}a\ne0\\a\ne-3\\a\ne3\end{cases}\)
Vậy để B xác định thì \(a\ne0\) và \(a\ne\pm3\)
b) \(B=\frac{\left(a+3\right)^2}{2a^2+6a}\cdot\left(1-\frac{6a-18}{a^2-9}\right)\)
\(=\frac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\frac{\left(a+3\right)\left(a-9\right)}{\left(a+3\right)\left(a-3\right)}\)
\(=\frac{a+3}{2a}\cdot\frac{a-9}{a+3}\)
\(=\frac{a-9}{2a}\)
NN
8 tháng 12 2017
a) ĐKXĐ: \(\left\{{}\begin{matrix}2a^2+6a\ne0\\a^2-9\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a\left(a+3\right)\ne0\\\left(a-3\right)\left(a+3\right)\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a\ne0\\a-3\ne0\\a+3\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a\ne0\\a\ne3\\a\ne-3\end{matrix}\right.\)
b) \(B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(\dfrac{a^2-9}{a^2-9}-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{\left(a^2-9\right)-\left(6a-18\right)}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{a^2-9-6a+18}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{a^2-6a+9}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{\left(a-3\right)^2}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{\left(a-3\right)^2}{\left(a-3\right)\left(a+3\right)}\)
\(\Leftrightarrow B=\dfrac{a+3}{2a}.\dfrac{a-3}{a+3}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)\left(a-3\right)}{2a\left(a+3\right)}\)
\(\Leftrightarrow B=\dfrac{a-3}{2a}\)
29 tháng 11 2016
\(A=\frac{\left[x\left(x^2-x+1\right)\right]-\left[\left(x+1\right)\left(3-3x\right)\right]+\left[x+4\right]}{x^3+1}\)
\(A=\frac{\left(x^3-x^2+x\right)+3\left(x^2-1\right)+\left(x+4\right)}{x^3+1}=\frac{x^3+2x^2+2x+1}{x^3+1}\)
\(A=\frac{\left(x^3+1\right)+2x\left(x+1\right)}{x^3+1}=1+\frac{2x}{x^2-x+1}\)
29 tháng 11 2016
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x\left(x^2-x+1\right)-\left(3+3x\right)\left(x+1\right)+\left(x+4\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-9x-3-3x^2+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2-3x^2+x-9x+x+3+4}{x^3+1}\)
\(A=\frac{x^3+2x^2-4x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{a^4-a^3+a-1}{a^4-a^3+2a^2-a+1}\)
\(=\frac{a^3\left(a-1\right)+\left(a-1\right)}{a^2\left(a^2-a+1\right)+\left(a^2-a+1\right)}\)
\(=\frac{\left(a-1\right)\left(a^3+1\right)}{\left(a^2-a+1\right)\left(a^2+1\right)}\)
\(=\frac{\left(a-1\right)\left(a+1\right)\left(a^2-a+1\right)}{\left(a^2-a+1\right)\left(a^2+1\right)}\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{\left(a^2+1\right)}=\frac{a^2-1}{a^2+1}=1-\frac{2}{a^2+1}\)
Vậy : \(\frac{a^4-a^3+a-1}{a^4-a^3+2a^2-a+1}\)\(=1-\frac{2}{a^2+1}\)