a) \(a\ne\frac{5}{2};\frac{2}{3}\)

Đặt \(A=\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2+\frac{2}{3a-2}-\frac{4}{2a-5}\)

\(A=2\Leftrightarrow\frac{2}{3a-2}-\frac{4}{2a-5}=0\Leftrightarrow4a-12a+8=0\)

\(\Leftrightarrow-8a-2=0\Leftrightarrow-2\left(4a+1\right)=0\Leftrightarrow a=-\frac{1}{4}\)

Vậy A=2 <=> a=-1/4

b) \(a\ne-\frac{4}{3};-4\)

Đặt \(B=\frac{3a+2}{3a+4}+\frac{a-2}{a+4}=2-\frac{2}{3a+4}-\frac{6}{a+4}\)

\(B=2\Leftrightarrow-\frac{2}{3a+4}-\frac{6}{a+4}=0\Leftrightarrow-2a-8-18a-24=0\)

\(\Leftrightarrow-20a-32=0\Leftrightarrow a=-\frac{8}{5}\)

Vậy B=2 <=> a= -8/5

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

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

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

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

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

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

\(\Leftrightarrow20a=-12\)

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

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

Ta có

\(\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}=\frac{3a^2+15ab-6b^2}{9a^2-b^2}\left(1\right)\)

Ta lại có

\(6a^2-15ab+5b^2=0\)

\(\Leftrightarrow9a^2-b^2=3a^2+15ab-6b^2\left(2\right)\)

Từ (1) và (2) => Q = 1

Đề thiếu rồi 

kết quả =1

     \(10a^2-b^2+ab=0\)

\(\Rightarrow10a^2+6ab-5ab-3b^2=0\)

\(\Rightarrow2a\left(5a+3b\right)-b\left(5a+3b\right)=0\)

\(\Rightarrow\left(5a+3b\right)\left(2a-b\right)=0\)

Mà \(b>a>0\Rightarrow5a+3b>0\)

Do đó: \(2a-b=0\Rightarrow2a=b\)

Ta có: \(B=\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}\)

             \(=0+\frac{10a-a}{3a+2a}\) (vì b = 2a)

              \(=0+\frac{9}{5}=\frac{9}{5}\)

Vậy \(A=\frac{9}{5}\)

Chúc bạn học tốt.

minh văn nguyễn