Rút gọn phân thức sau :
\(\frac{a^4-3a^2+1}{a^4-a^2-2a-1}\)
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HN
25 tháng 11 2018
Ta có : \(\frac{a^4-3a^2+1}{a^4-a^2-2a-1}\) \(=\frac{\left(a^4-2a^2+1\right)-a^2}{\left(a^4-a^3-a^2\right)+\left(a^3-a^2-a\right)+\left(a^2-a-1\right)}\)
\(=\frac{\left(a^2-1\right)^2-a^2}{a^2\left(a^2-a-1\right)+a\left(a^2-a-1\right)+\left(a^2-a-1\right)}\)
\(=\frac{\left(a^2-a-1\right)\left(a^2+a-1\right)}{\left(a^2-a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{a^2+a-1}{a^2+a+1}\)
29 tháng 11 2016
1, b) \(\frac{x^2+y^2-4+2xy}{x^2-y^2+4+4x}\) = \(\frac{\left(x^2+2xy+y^2\right)-4}{\left(x^2+4x+4\right)-y^2}\) =\(\frac{\left(x+y\right)^2-2^2}{\left(x+2\right)^2-y^2}\)= \(\frac{\left(x+y+2\right)\left(x+y-2\right)}{\left(x+2+y\right)\left(x+2-y\right)}\) = \(\frac{x+y-2}{x+2-y}\)
2, A= \(\frac{a^2+ax+ab+bx}{a^2+ax-ab-bx}\) = \(\frac{\left(a^2+ax\right)+\left(ab+bx\right)}{\left(a^2+ax\right)-\left(ab+bx\right)}\) = \(\frac{a\left(a+x\right)+b\left(a+x\right)}{a\left(a+x\right)-b\left(a+x\right)}\)= \(\frac{\left(a+x\right)\left(a+b\right)}{\left(a+x\right)\left(a-b\right)}\)= \(\frac{a+b}{a-b}\)
LM
19 tháng 7 2017
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
2 tháng 7 2021
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
NV
0
\(\frac{a^4-3a^2+1}{a^4-a^2-2a-1}\)
Theo đề bài ta có :
Tử số : \(a^4-2a^2+1-a^2\)
\(=\left(a^2-1\right)^2-a^2\)
\(=\left(a^2-1+a\right)\left(a^2-1-a\right)\)
Mẫu số : \(a^4-\left(a^2+2a+1\right)\)
\(=a^4-\left(a+1\right)^2\)
\(=\left(a^2+a+1\right)\left(a^2-a-1\right)\)
Phân thức bằng \(\frac{a^2+a-1}{a^2+a+1}\)với điều kiện \(a^2-a-1\ne0\)