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DL
1
DN
12 tháng 4 2016
P=\(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{\left(a^3-4a^2\right)-\left(a-4\right)}{\left(a^3-8\right)-\left(7a^2-14a\right)}\)
\(=\frac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(=\frac{\left(a-4\right)\left(a^2-1\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(=\frac{\left(a-4\right)\left(a^2-1\right)}{\left(a-2\right)\left(\left(a^2-4a\right)-\left(a-4\right)\right)}\)
\(=\frac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a\left(a-4\right)-\left(a-4\right)\right)}\)
\(=\frac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a-4\right)\left(a-1\right)}\)
\(=\frac{a+1}{a-2}\)
Chúc bạn học giỏi, k cho mình nhé!!!
HK
1
14 tháng 8 2016
\(\frac{a^3-4a^2-a+a}{a^3-7a^2+14a-8}=\frac{a^3-4a^2}{a^3-4a^2-3a^2+12a+2a-8}\)
\(=\frac{a^2\left(a-4\right)}{a^2\left(a-4\right)-3a\left(a-4\right)+2\left(a-4\right)}=\frac{a^2\left(a-4\right)}{\left(a-4\right)\left(a^2-3a+2\right)}\)
\(=\frac{a^2}{a^2-3a+2}=\frac{a^2}{a\left(a-2\right)-\left(a-2\right)}=\frac{a^2}{\left(a-2\right)\left(a-1\right)}\)
Ủng hộ mik nhé!!!!
T
1
HC
13 tháng 9 2016
ta có : \(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}\)
= \(\frac{\left(a-1\right)\left(a+1\right)\left(a-4\right)}{\left(a-4\right)\left(a-2\right)\left(a-1\right)}\)
= \(\frac{a+1}{a-2}\)
nhớ nha
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 3 2019
ĐKXĐ: \(a\ne\pm1;2;4\)
\(P=\frac{a^3-5a^2+4a+a^2-5a+4}{a^3-5a^2+4a-2a^2+10a-8}=\frac{a\left(a^2-5a+4\right)+\left(a^2-5a+4\right)}{a\left(a^2-5a+4\right)-2\left(a^2-5a+4\right)}\)
\(P=\frac{\left(a+1\right)\left(a^2-5a+4\right)}{\left(a-2\right)\left(a^2-5a+4\right)}=\frac{a+1}{a-2}\)
b/ \(P=\frac{a+1}{a-2}=1+\frac{3}{a-2}\)
\(P\) nguyên khi \(a-2=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(a-2=-3\Rightarrow a=-1\left(l\right)\)
\(a-2=-1\Rightarrow a=1\left(l\right)\)
\(a-2=1\Rightarrow a=3\)
\(a-2=3\Rightarrow a=5\)
Vậy \(\left[{}\begin{matrix}a=3\\a=5\end{matrix}\right.\) thì P nguyên
9 tháng 3 2019
\(P=\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{\left(a-4\right)\left(a+1\right)\left(a-1\right)}{\left(a-1\right)\left(a-2\right)\left(a-4\right)}=\frac{a+1}{a-2}\)
b \(P=\frac{a-2+3}{a-2}=1+\frac{3}{a-2}\)
Để P nhận giá trị nguyên \(\left(a-2\right)\inƯ\left(3\right)=\left\{1;-1;-3;3\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}a-2=1\\a-2=-1\\a-2=3\\a-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\\a=1\\a=5\\a=-1\end{matrix}\right.\)
28 tháng 2 2022
a: \(VT=\dfrac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(=\dfrac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a^2-5a+4\right)}\)
\(=\dfrac{\left(a-4\right)\left(a+1\right)}{\left(a-4\right)\left(a-1\right)}=\dfrac{a+1}{a-1}=VP\)
b: \(VT=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2}{x^2+1}=VP\)
KB
1
HV
14 tháng 11 2017
Có \(\text{VT }=\) \(\dfrac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}\)
\(\Rightarrow VT=\dfrac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(\Rightarrow VT=\dfrac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(\Rightarrow VT=\dfrac{\left(a+1\right)\left(a^2-5a+4\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(\Rightarrow\dfrac{a+1}{a-2}\)
\(\Rightarrow VT=VP\)
\(\Rightarrowđpcm\)
LM
19 tháng 7 2017
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)