K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

6 tháng 7 2016

Tử = \(a^3-3a+2=a^3-1-3a+3\)

                                \(=\left(a-1\right)\left(a^2+a+1\right)-3\left(a-1\right)\)

                                \(=\left(a-1\right)\left(a^2+a-2\right)\)

                                \(=\left(a-1\right)\left(a-1\right)\left(a+2\right)=\left(a-1\right)^2\left(a+2\right)\)

Mẫu =\(2a^3-7a^2+8a-3=2a\left(a^2-2a+1\right)-3\left(a^2-2a+1\right)\)

                                                 \(=\left(a-1\right)^2\left(2a-3\right)\)

=>\(\frac{a^3-3a+2}{2a^3-7a^2+8a-3}=\frac{\left(a-1\right)^2\left(a+2\right)}{\left(a-1\right)^2\left(2a-3\right)}=\frac{a+2}{2a-3}\)

Nhớ h cho mik nhé

\(\dfrac{a^3-3a+2}{2a^3-7a^2+8a-3}\)

\(=\dfrac{a^3-a-2a+2}{2a^3-2a^2-5a^2+5a+3a-3}\)

\(=\dfrac{a\left(a-1\right)\left(a+1\right)-2\left(a-1\right)}{2a^2\left(a-1\right)-5a\left(a-1\right)+3\left(a-1\right)}\)

\(=\dfrac{\left(a-1\right)\left(a^2+a-2\right)}{\left(a-1\right)\left(2a^2-5a+3\right)}\)

\(=\dfrac{\left(a+2\right)\left(a-1\right)}{\left(a-1\right)\left(2a-3\right)}\)

\(=\dfrac{a+2}{2a-3}\)

19 tháng 7 2017

c)\(P=\)\(\frac{\left(a-b\right)^2-c^2}{\left(a-b+c\right)^2}=\frac{\left(a-b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)^2}=\frac{a-b-c}{a-b+c}\)

19 tháng 7 2017

b)\(M\)\(=\frac{\left(a+2\right)\left(a-1\right)^2}{\left(2a-3\right)\left(a-1\right)^2}=\frac{a+2}{2a-3}\)

19 tháng 7 2017

a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)

\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)

\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)

\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)

\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)

\(=\left(a^2-a+2\right)\left(a+2\right)\)

\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)

17 tháng 5 2020

em chịu

\(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)

\(=21a^2-35a+8a^2+2a-12a-3-36a^2+24a-4\)

\(=-7a^2+4a-7\)

4 tháng 11 2019

ĐKXĐ: \(a\ne\frac{3}{2},a\ne-\frac{3}{2}\)

a, \(P=\left(\frac{a-1}{2a-3}-\frac{3a}{4a+6}+\frac{7a-2a^2-1}{18-8a^2}\right):\frac{1}{6-4a}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}+\frac{7a-2a^2-1}{2\left(9-4a^2\right)}\right):\frac{-1}{4a-6}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}-\frac{7a-2a^2-1}{2\left(4a^2-9\right)}\right):\frac{-1}{2\left(2a-3\right)}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}-\frac{7a-2a^2-1}{2\left(2a-3\right)\left(2a+3\right)}\right)\left[-2\left(2a-3\right)\right]\)

\(=\left[\frac{2\left(a-1\right)\left(2a+3\right)-3a\left(2a-3\right)-\left(7a-2a^2-1\right)}{2\left(2a-3\right)\left(2a+3\right)}\right]\left[-2\left(2a-3\right)\right]\)

\(=\frac{4a-5}{2\left(2a-3\right)\left(2a+3\right)}\left[-2\left(2a-3\right)\right]\)

\(=-\frac{\left(4a-5\right)}{2a+3}=\frac{5-4a}{2a+3}\)