Question

помогите решить пж
помогите решить пж
помогите решить пж ​

Answer 1

$2. \ \dfrac{x+1}{x-3}-\dfrac{9}{x+3}=\dfrac{24}{x^{2}-9}, \ x\neq \{-3,3\}\\\\\dfrac{(x+1)(x+3)}{(x-3)(x+3)}-\dfrac{9(x-3)}{(x-3)(x+3)}-\dfrac{24}{x^{2}-9}=0\\\\\dfrac{(x+1)(x+3)-9(x-3)-24}{x^{2}-9}=0\\\\x^{2}+3x+x+3-9x+27-24=0\\\\x^{2}-5x+6=0\\\\x^{2}-2x-3x+6=0\\\\x(x-2)-3(x-2)=0\\\\(x-3)(x-2)=0\\\\x=2\ \lor\ x=3 \ \land\ x\neq \{-3,3\}\\\\\boxed{x=2}$

$3. \ \dfrac{x+1}{x-1}-\dfrac{4}{x^{2}-1}-\dfrac{x-1}{x+1}=0, \ x\neq \{-1,1\}\\\\\dfrac{(x+1)(x+1)}{(x-1)(x+1)}-\dfrac{4}{x^{2}-1}-\dfrac{(x-1)(x-1)}{(x-1)(x+1)}=0\\\\\dfrac{(x+1)(x+1)-4-(x-1)(x-1)}{x^{2}-1}=0\\\\(x+1)^{2}-4-(x-1)^{2}=0\\\\x^{2}+2x+1-4-(x^{2}-2x+1)=0\\\\x^{2}+2x-3-x^{2}+2x-1=0\\\\4x-4=0\ \ \ \ \ |:4\\\\x-1=0\\\\x=1\ \land \ x\neq \{-1,1\}\\\\x\in\emptyset$

$4. \ \dfrac{x-3}{x+5}-\dfrac{x-9}{x-1}=\dfrac{48}{(x+5)(x-1)}, \ x\neq \{-5,1\}\\\\\dfrac{(x-1)(x-3)}{(x-1)(x+5)}-\dfrac{(x+5)(x-9)}{(x-1)(x+5)}-\dfrac{48}{(x-1)(x+5)}=0\\\\\dfrac{(x-1)(x-3)-(x+5)(x-9)-48}{(x-1)(x+5)}=0\\\\x^{2}-3x-x+3-(x^{2}-9x+5x-45)-48=0\\\\x^{2}-4x+3-x^{2}+4x+45-48=0\\\\48-48=0\\\\0=0\\\\\boxed{x\in\mathbb{R}/\{-5,1\}}$