a. Với a = -3 ta được:

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}+\dfrac{27-3}{x^2-9}=0\)

\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{24}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9+24=0\)

\(\Leftrightarrow12x+24=0\)

\(\Leftrightarrow x=-2\)

Giải phương trình :

\(\dfrac{x-a}{x+a}-\dfrac{x+a}{x-a}+\dfrac{3a^2+a}{x^2-a^2}=0\)

a) Với a = -3

\(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}+\dfrac{27+3}{x^2-3^2}=0\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)

Ta có : \(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}+\dfrac{27+3}{x^2-3^2}\)

\(\Leftrightarrow\) \(\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{27+3}{\left(x+3\right)\left(x-3\right)}=0\)

Khử mẫu ta có : \(\left(x-3\right)^2-\left(x+3\right)^2+27+3=0\)

⇔ \(x^2+6x+9-x^2+6x-9+30=0\)

\(\Leftrightarrow12x+30=0\)

\(\Leftrightarrow12x=-30\)

\(\Leftrightarrow x=-\dfrac{5}{2}\)

Tập nghiệm của pt là: \(S=\left\{-\dfrac{5}{2}\right\}\)

b) Với a = 1

\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{3+3}{x^2-1}=0\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)

Ta có : \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{3+3}{x^2-1}=0\)

\(\Leftrightarrow\) \(\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3+3}{\left(x+1\right)\left(x-1\right)}=0\)

Khử mẫu ta có : \(\left(x-1\right)^2-\left(x+1\right)^2+6=0\)

\(\Leftrightarrow x^2+x-1-x^2+x+1+6=0\)

\(\Leftrightarrow2x+6=0\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

Tập nghiệm của pt là : \(S=\left\{-3\right\}\)

ĐKXĐ: x khác a

b. Với a = 1 ta có pt:

\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{3+1}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow x^2-2x+2-x^2-2x-2+4=0\)

\(\Leftrightarrow-4x+4=0\)

\(\Leftrightarrow-4x=-4\)

\(\Leftrightarrow x=1\)( loại)

Vậy pt vô nghiệm với a = 1

Đk:\(a\ne\pm x\)

Pt \(\Leftrightarrow\dfrac{\left(a+x\right)^2-\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+x\right)}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)

\(\Leftrightarrow\dfrac{2\left(a^2+x^2\right)}{a^2-x^2}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)

\(\Leftrightarrow2a^2+2x^2=3a^2+a\)

\(\Leftrightarrow a^2+a-2x^2=0\) (1)

Thay \(x=\dfrac{1}{2}\) vào (1) ta được:

\(a^2+a-2\left(\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow a^2+a-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{-1+\sqrt{3}}{2}\\a=\dfrac{-1-\sqrt{3}}{2}\end{matrix}\right.\) (tm)

Vậy...

a: Khi a=-3 thì phương trình sẽ là:

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}+\dfrac{3\cdot9-3}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9+24=0\)

=>12x=-24

hay x=-2

b: Khi a=1 thì phương trình trở thành:

\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x^2-2x+1-x^2-2x-1+4=0\)

=>-4x+4=0

hay x=1(loại)

Bên h mik đã từng giải rồi ạ :33

https://h.vn/hoi-dap/question/940816.html

a) Thay a = -1 vào phương trình

\(\dfrac{x-1}{x+3}+\dfrac{x-3}{x+1}=2\)

\(\Rightarrow\dfrac{x^2-1+x^2-9}{\left(x+3\right)\left(x+1\right)}=2\)

\(\Rightarrow2x^2-10=2\left(x+3\right)\left(x+1\right)=2x^2+8x+6\)

\(\Rightarrow2x^2+8x+6-2x^{10}+10=0\)

\(\Rightarrow8x+16=0\Rightarrow x=-2\)

b, c Làm tương tự như câu a

d)

Phương trình nhận x = 1 làm nghiệm

=> \(\dfrac{1+a}{1+3}+\dfrac{1-3}{1-a}=2\)

\(\Rightarrow\dfrac{a+1}{4}+\dfrac{2}{a-1}=2\)

\(\Rightarrow\dfrac{a^2-1+8}{4\left(a-1\right)}=2\)

\(\Rightarrow a^2+7=2\left(4a-1\right)=8a-2\)

\(\Rightarrow a^2-8x+9=0\)

\(\Rightarrow\left[{}\begin{matrix}a=4+\sqrt{7}\\a=4-\sqrt{7}\end{matrix}\right.\)

\(\Leftrightarrow A=\dfrac{\left(x-a\right)^2-\left(x+a\right)^2+3a^2+a}{\left(x-a\right)\left(x+a\right)}\)

\(\Leftrightarrow A=\dfrac{-4ax+3a^2+a}{\left(x-a\right)\left(x+a\right)}\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|\ne a\\4ax=a\left(3a+1\right)\left(1\right)\end{matrix}\right.\)

a) với a=-3

\(\left(1\right)\Leftrightarrow4x=3.\left(-3\right)+1\Rightarrow x=-2\)(NHAN)

b)với a=-1

\(\left(1\right)\Leftrightarrow4x=3.\left(-1\right)+1\Rightarrow x=-\dfrac{2}{4}=-\dfrac{1}{2}\)(NHẬN)

c)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\x=\dfrac{3a+1}{4}=0,5\Rightarrow a=\dfrac{1}{3}\left(nhan\right)\end{matrix}\right.\)