a) \(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)

<=> (2a - 9)(3a - 2) + 3a(2a - 5) = 2(2a - 5)(3a - 2)

<=> 6a² - 4a - 27a + 16 + 6a² - 15a = 12a² - 8a - 30a + 20

<=> 12a² - 44a + 16 = 12a² - 38a + 20

<=> 12a² - 44a + 16 - 12a² = -38a + 20

<=> -44a + 16 = -36a + 20

<=> -44a + 16 + 36a = 20

<=> -8a + 16 = 20

<=> -8a = 20 - 16

<=> -8a = 4

<=> a = -4/8 = -1/2

b) nhân chéo và làm tương tự

Biểu thức nguyên khi 3a-5 chia hết cho 2a-9

=> 2(3a-5) chia hết cho 2a-9 

2(3a-5)=6a-10=6a-27+17=3(2a-9)+17

=> 3a-5 chia hết cho 2a-9 khi 17 chia hết cho 2a-9. Có các TH:

+/ 2a-9=1 => a=10/2=5

+/ 2a-9=-1 => a=8/2=4

+/ 2a-9=17 => a=26/2=13

+/ 2a-9=-17 => a=-8/2=-4

ĐS: a={-4; 4; 5; 13}

ta thấy rằng 5 phải chia hết cho a tức là 

a(U)5=1,-1;5,-5

vậy a 1,-1,5,-5 thì x có giá trị nguyên 

a=-4;4;13

a) \(\frac{2a^2-3a-2}{a^2-4}=2\)

\(\Rightarrow2a^2-3a-2=2\left(a^2-4\right)\)

\(\Rightarrow2a^2-3a-2=2a^2-4\)

\(\Rightarrow-3a-2=-4\)

\(\Rightarrow-3a=-2\Rightarrow a=\frac{2}{3}\)

b) \(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)

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

\(\Rightarrow\frac{6a^2-6}{3a^2+10a+3}=2\)

\(\Rightarrow6a^2-6=2\left(3a^2+10a+3\right)\)

\(\Rightarrow6a^2-6=6a^2+20a+6\)

\(\Rightarrow-6=20a+6\Rightarrow20a=-12\)

\(\Rightarrow a=\frac{-3}{5}\)

a, \(\frac{2a^2-3a-2}{a^2-4}=2\)

\(\Leftrightarrow\frac{a\left(2a+1\right)-2\left(2a+1\right)}{a^2-4}=2\)

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

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

\(\Leftrightarrow\frac{2a+1}{a+2}=2\)

\(\Leftrightarrow2a+1=2\left(a+2\right)\Leftrightarrow2a+1=2a+4\Leftrightarrow2a+1-2a-4=0\)

\(\Leftrightarrow-3\ne0\)(voli)

b, \(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)

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

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

\(\Leftrightarrow6a^2-6=6a^2+20a+6\)

\(\Leftrightarrow6a^2-6-6a^2-20a-6=0\)

\(\Leftrightarrow-12-20a=0\)

\(\Leftrightarrow20a=-12\)

\(\Leftrightarrow a=-\frac{3}{5}\)