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22 tháng 5 2022
\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{a^3-1}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)
17 tháng 4 2021
Ta có:
\(VT=\left[\dfrac{16a-a^2-\left(3+2a\right)\left(a+2\right)-\left(2-3a\right)\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}\right]:\dfrac{a-1}{a^3+4a^2+4a}\)
\(=\dfrac{16a-a^2-3a-6-2a^2-4a-2a+4+3a^2-6a}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}\)
\(=\dfrac{a-2}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}=\dfrac{a\left(a+2\right)}{a-1}\left(a\ne\pm2;a\ne1\right)\)
\(=a-\dfrac{a\left(a+2\right)}{a-1}=\dfrac{a^2-a-a^2-2a}{-1}=\dfrac{-3a}{a-1}=\dfrac{3a}{1-a}=VP\left(đpcm\right)\)
AH
Akai Haruma
Giáo viên
4 tháng 1 2021
Lời giải:
a) \(\frac{3a+1}{ab}+\frac{2a-1}{ab}=\frac{3a+1+2a-1}{ab}=\frac{5a}{ab}=\frac{5}{b}\)
b)
\(\frac{3}{2(y-1)}+\frac{7}{y}=\frac{3y+14(y-1)}{2y(y-1)}=\frac{17y-14}{2y(y-1)}\)
Y
10 tháng 7 2023
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\left(dk:a\ne1,a\ne-1\right)\)
\(=-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right):\dfrac{2a+1}{\left(a-1\right)\left(a+1\right)}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}.\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)
\(=-\dfrac{2}{2a+1}\)
H
11 tháng 7 2023
\(-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}-\dfrac{3a+1}{1-a^2}\right):\dfrac{2a+1}{a^2-1}\\ =-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a+1}{a^2-1}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{a-1}{a+1}-\dfrac{a}{a-1}+\dfrac{3a-1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{\left(a-1\right)^2}{\left(a+1\right)\left(a-1\right)}-\dfrac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}+\dfrac{3a+1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{\left(a-1\right)^2-a\left(a+1\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\)\(=-\left(\dfrac{a^2-2a+1-\left(a^2+a\right)+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{a^2-2a+1-a^2-a+3a+1}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =-\left(\dfrac{2}{\left(a-1\right)\left(a+1\right)}\right).\dfrac{\left(a-1\right)\left(a+1\right)}{2a+1}\\ =\dfrac{-2.\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a+1\right).\left(2a+1\right)}\\ =-\dfrac{2}{2a+1}\)
__
\(-\dfrac{2}{2a+1}=\dfrac{3}{a-1}\\ \Leftrightarrow-2\left(a-1\right)=3\left(2a+1\right)\\ \Leftrightarrow-2a+2-6a-3=0\\ \Leftrightarrow-8a-1=0\\ \Leftrightarrow-8a=1\\ \Leftrightarrow a=-\dfrac{1}{8}\)
13 tháng 1 2018
minh giai phan d, nha bn :
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c
\(P=\dfrac{\left(2a-1\right)\left(3a+1\right)+\left(5-a\right)\left(3a-1\right)}{\left(3a-1\right)\left(3a+1\right)}\)
\(=\dfrac{6a^2-a-1-3a^2+16a-5}{9a^2-1}=\dfrac{3a^2+15a-6}{9a^2-1}\)
\(=\dfrac{\left(30a^2+15a\right)-\left(27a^2-3\right)-9}{9a^2-1}\)
\(=\dfrac{3\left(10a^2+5a\right)-9-3\left(9a^2-1\right)}{9a^2-1}\)
\(=\dfrac{3.3-9-3\left(9a^2-1\right)}{9a^2-1}=-3\) (vì 10a2 + 5a = 3)