a.
\(\dfrac{2a^2-3a-2}{a^2-4}=2\)
\(\Leftrightarrow\dfrac{2a^2-4a+a-2}{\left(a-2\right)\left(a+2\right)}=2\)
\(\Leftrightarrow\dfrac{\left(2a^2-4a\right)+\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=2\)
\(\Leftrightarrow\dfrac{2a\left(a-2\right)+\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=2\)
\(\Leftrightarrow\dfrac{\left(2a+1\right)\left(a-2\right)}{\left(a-2\right)\left(a+1\right)}=2\)
\(\Leftrightarrow\dfrac{2a+1}{a+1}=2\)
\(\Leftrightarrow\dfrac{2a+1}{a+1}=\dfrac{2\left(a+1\right)}{a+1}\)
\(\Leftrightarrow2a+1=2a+2\)
Suy ra pt vô nghiệm
a) \(\dfrac{2a^{2^{ }}-3a-2}{a^2-4}\)=2
<=> \(\dfrac{2a^{2^{ }}-3a-2}{\left(a-2\right)\left(a+2\right)}\)=2 (1)
ĐKXĐ: a-2 #0 => a#2
a+2#0 -> a#-2
(1) <=> \(\dfrac{2a^{2^{ }}-3a-2}{\left(a-2\right)\left(a+2\right)}\)= \(\dfrac{2\left(a^{^2}-4\right)}{\left(a-2\right)\left(a+2\right)}\)
=> 2a2 - 3a - 2 = 2a2 - 8
<=> 2a2 - 3a - 2 - 2a2 + 8 = 0
<=> -3a + 6 = 0
<=> -3 ( a-2)
<=> -3 = 0 ( vô no )
a-2 = 0 => a = 2
Vậy với A=2 thì biểu thức có giá trị = 2
á t bị nhầm chút:
\(\Leftrightarrow\dfrac{2a+1}{a+2}=2\)
\(\Leftrightarrow\dfrac{2a+1}{a+2}=\dfrac{2\left(a+2\right)}{a+2}\)
\(\Leftrightarrow2a+1=2a+4\)
suy ra pt vô nghiệm
a) \(\dfrac{2a^2-3a-2}{a^2-4}=2\left(ĐKXĐ:a\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{2a^2-3a-2}{a^2-4}=\dfrac{2\left(a^2-4\right)}{a^2-4}\)
\(\Rightarrow2a^2-3a-2=2a^2-8\)
\(\Leftrightarrow2a^2-2a^2-3a=-8+2\)
\(\Leftrightarrow-3a=-6\)
\(\Leftrightarrow a=2\) (không thỏa mãn ĐKXĐ)
Vậy S = \(\varnothing\)
b) \(\dfrac{3a-1}{3a+1}+\dfrac{a-3}{a+3}=2\left(ĐKXĐ:a\ne\dfrac{-1}{3};a\ne-3\right)\)
\(\Leftrightarrow\dfrac{\left(3a-1\right)\left(a+3\right)+\left(a-3\right)\left(3a+1\right)}{\left(3a+1\right)\left(a+3\right)}=\dfrac{2\left(3a+1\right)\left(a+3\right)}{\left(3a+1\right)\left(a+3\right)}\)
\(\Rightarrow\left(3a-1\right)\left(a+3\right)+\left(a-3\right)\left(3a+1\right)=2\left(3a+1\right)\left(a+3\right)\)
\(\Leftrightarrow3a^2+9a-a-3+3a^2+a-9a-3=\left(6a+2\right)\left(a+3\right)\)
\(\Leftrightarrow6a^2-6=6a^2+18a+2a+6\)
\(\Leftrightarrow6a^2-6a^2-20a=12\)
\(\Leftrightarrow-20a=12\)
\(\Leftrightarrow a=\dfrac{-3}{5}\) (thỏa mãn ĐKXĐ)
Vậy .................
b) \(\dfrac{3a-1}{3a+1}\)+ \(\dfrac{a-3}{a+3}\)=2 (1)
ĐKXĐ : 3a +1 # 0 => a# -1/3
a+3 # 0 => a # -3
(1)<=> \(\dfrac{\left(3a-1\right)\left(a+3\right)}{MTC}\)+ \(\dfrac{\left(a-3\right)\left(3a+1\right)}{MTC}\)= \(\dfrac{2\left(a-3\right)\left(3a+1\right)}{MTC}\)
=> 3a2 + 9a - a - 3 + 3a2 + a - 9a - 3 = 2 ( 3a2 + a - 9a - 3)
<=> 3a2 + 9a - a - 3 + 3a2 + a - 9a - 3 - 6a2 - 2a + 18a +6 = 0
<=> 16a = 0
<=> a=0
Vậy với a = 0 thì biểu thức có giá trị = 2
a) \(\dfrac{2a^2-3a-2}{a^2-4}=2\)
ĐKXĐ: \(a\ne2;a\ne-2\)
\(\Leftrightarrow2a^2-3a-2=2a^2-8\)
\(\Leftrightarrow3a=6\)
\(\Leftrightarrow a=2\) (l)
Vậy S=\(\varnothing\)
b) \(\dfrac{3a-1}{3a+1}+\dfrac{a-3}{a+3}=2\)
\(ĐKXĐ:a\ne\dfrac{-1}{3};a\ne-3\)
\(\Leftrightarrow\dfrac{\left(3a-1\right)\left(a+3\right)+\left(a-3\right)\left(3a+1\right)}{\left(a+3\right)\left(3a+1\right)}=2\)
\(\Leftrightarrow3a^2-a+9a-3+3a^2+a-9a-3=2\left(3a^2+10a+3\right)\)
\(\Leftrightarrow6a^2-6=6a^2+20a+6\)
\(\Leftrightarrow20a=-12\)
\(\Leftrightarrow a=\dfrac{-5}{3}\left(n\right)\)
Vậy S=\(\left\{-\dfrac{5}{3}\right\}\)
a, \(\dfrac{2a^2-3a-2}{a^2-4}=2\) ĐKXĐ: \(a\ne2,a\ne-2\)
\(\Rightarrow2a^2-3a-2=2\left(a^2-4\right)\)
\(\Leftrightarrow2a^2-3a-2=2a^2-8\)
\(\Leftrightarrow2a^2-3a-2a^2=-8+2\)
\(\Leftrightarrow-3a=-6\)
\(\Leftrightarrow a=2\) (loại)
Vậy không có giá trị nào của a để biểu thức có giá trị bằng 2.
b, \(\dfrac{3a-1}{3a+1}+\dfrac{a-3}{a+3}=2\) ĐKXĐ: \(a\ne\dfrac{-1}{3};a\ne-3\)
\(\Rightarrow\left(3a-1\right)\left(a+3\right)+\left(a-3\right)\left(3a+1\right)=2\left(a+3\right)\left(3a+1\right)\)
\(\Leftrightarrow3a^2+9a-a-3+3a^2+a-9a-3=6a^2+2a+18a+6\)
\(\Leftrightarrow3a^2+9a-a+3a^2+a-9a-6a^2-2a-18a=6+3+3\)
\(\Leftrightarrow-20a=12\)
\(\Leftrightarrow a=\dfrac{-3}{5}\) (TĐK)
Vậy a = 2 thì biểu thức có giá trị bằng 2.