Rút gọn phân thức: N = x 7 + x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 x 2 − 1 .
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Cao Minh Tâm
25 tháng 11 2018
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VT
1
VI
2
13 tháng 6 2016
\(\frac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\left(DK:x\ne-1;x\ne1\right)\)
\(=\frac{x^4\left(x^3+x^2+x+1\right)+\left(x^3+x^2+x+1\right)}{x^2-1}\)
\(=\frac{x^4\left[x\left(x^2+1\right)+x^2+1\right]+\left[x\left(x^2+1\right)+x^2+1\right]}{x^2-1}\)
\(=\frac{\left(x^4+1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x^2+1\right)\left(x^4+1\right)}{x-1}\)
13 tháng 6 2016
\(\frac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\)
\(=\frac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^6+x^4+x^2\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^6+x^4+x^2}{x+1}\)
\(=\frac{x^2\left(x^3+x^2+1\right)}{x+1}\)
5 tháng 12 2017
\(\frac{x^{10}-x^8-x^7+x^6+x^6+x^4-x^3-x^2+1}{x^{30}+x^{24}+x^{18}+x^{12}+x^6+1}=\frac{(x^{10}-x^8+x^6)-(x^7-x^5+x^3)+(x^4-x^2+1)}{ (x^{30}+x^{18}+x^{24})+(x^{12}+x^6+1)} \)
=\(\frac{(x^4-x^2+1)(x^6-x^3+1)}{(x^{12}+x^6+1)(x^{18}+1 )}=\frac{(x^4-x^2+1)(x^6-x^3+1)}{(x^{12}+2x^6+1-x^6) (x^6+1)(x^{12}-x^6+1)}=\frac{(x^4-x^2+1)(x^6-x^3+1)}{ (x^6-x^3+1)(x^6+x^3+1)(x^2+1)(x^4-x^2+1)(x^12-x^6+1 )} \)
=\(\frac{1}{(x^6+x^2+1)(x^2+1)(x^{12}-x^6+1)}\)
30 tháng 10 2023
a) ĐKXĐ:
\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\)
b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)
\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)
c) Thay x = - 1 vào A ta có:
\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)
17 tháng 8 2021
\(\dfrac{6x-x^2-5}{5x^6-x^7}\\ =\dfrac{-x^2+5x+x-5}{x^6\left(5-x\right)}\\ =\dfrac{\left(-x^2+5x\right)+\left(x-5\right)}{x^6\left(5-x\right)}\\ =\dfrac{-x\left(x-5\right)+\left(x-5\right)}{-x^6\left(x-5\right)}\\ =\dfrac{\left(-x+1\right)\left(x-5\right)}{-x^6\left(x-5\right)}\\ =\dfrac{-x+1}{-x^6}\)
17 tháng 8 2021
\(\dfrac{-x^2+6x-5}{5x^6-x^7}\)
\(=\dfrac{x^2-6x+5}{x^7-5x^6}\)
\(=\dfrac{\left(x-1\right)\left(x-5\right)}{x^6\left(x-5\right)}\)
\(=\dfrac{x-1}{x^6}\)
HT
0
T
11 tháng 11 2023
\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)
\(=\dfrac{\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]+1}{x^2+7x+11}\)
\(=\dfrac{\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1}{x^2+7x+11}\)
Đặt \(x^2+7x+11=y\), thay vào \(N\) ta được:
\(N=\dfrac{\left(y-1\right)\left(y+1\right)+1}{y}\)
\(=\dfrac{y^2-1+1}{y}\)
\(=\dfrac{y^2}{y}\)
\(=y\)
\(=x^2+7x+11\)
Vậy \(N=x^2+7x+11\).
\(\text{#}Toru\)
