Tìm các giá trị của a sao cho mỗi biểu thức sau có giá trị = 2 b) $\frac{3a-1}{3a+1}$ + $\frac{a-3}{a+3}$ c) $\frac{10}{3}$ - $\frac{3a-1}{4a+12}$ - $\frac{7a+2}{6a+18}$ d) $\frac{2a-9}{2a-5}$ + $\frac{3a}{3a-2}$

Đáp án + Giải thích các bước giải:

`b) (3a-1)/(3a+1) + (a-3)/(a+3)=2`

ĐK `a\ne-1/3; a\ne-3`

`<=> ((3a-1)(a+3))/((3a+1)(a+3)) + ((3a+1)(a-3))/((3a+1)(a+3))=(2(3a+1)(a+3))/((3a+1)(a+3))`

`=> (3a-1)(a+3) + (3a+1)(a-3)=2(3a+1)(a+3)`

`<=> 3a^2+9a-a-3 + 3a^2-9a+a-3=2(3a^2+9a+a+3)`

`<=> 6a^2-6=6a^2+20a+6`

`<=> 20a+6+6=0`

`<=> 20a+12=0`

`<=> a=-3/5 (tm)`

Vậy `S={-3/5}`

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`c) 10/3 - (3a-1)/(4a+12) - (7a+2)/(6a+18)=2`

ĐK: `a\ne-3`

`<=> 10/3 - (3a-1)/(4(a+3)) - (7a+2)/(6(a+3))=2`

`<=> (40(a+3))/(12(a+3)) - (3(3a-1))/(12(a+3)) - (2(7a+2))/(12(a+3))=(24(a+3))/(12(a+3))`

`<=> 40(a+3) - 3(3a-1) - 2(7a+2)=24(a+3)`

`<=> 40a+120 - 9a+3 - 14a-4=24a+72`

`<=> 40a+120 - 9a+3 - 14a-4-24a-72=0`

`<=> -7a+47=0`

`<=> a = 47/7 (tm)`

Vậy `S={47/7}`

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`d) (2a-9)/(2a-5)+(3a)/(3a-2)=2`

ĐK: `a\ne5/2; a\ne2/3`

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

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

`<=> 6a^2-4a-27a+18+6a^2-15a=2(6a^2-4a-15a+10)`

`<=> 12a^2-46a+18=2(6a^2-19a+10)`

`<=> 12a^2-46a+18=12a^2-38a+20`

`<=> -8a-2=0`

`<=> a=-1/4 (tm)`

Vậy `S={-1/4}`