b) ĐKXĐ : \(x\ne\pm1\)

\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)

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

c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-1+2x-\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{2}{\sqrt{x}}\)

a) ĐKXĐ : \(x\ge0;x\ne16\)

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)

\(=\left(\dfrac{\sqrt{x}.\left(\sqrt{x}-4\right)}{x-4}+\dfrac{4.\left(\sqrt{x}+4\right)}{x-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\left(\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\right).\dfrac{\sqrt{x}+2}{x+16}\)

\(=\dfrac{x+16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+16}\)

\(=\dfrac{1}{\sqrt{x}-2}\)

\(A=\dfrac{2}{x^2+2x}+\dfrac{2}{x^2+6x+8}+\dfrac{2}{x^2+10x+24}+\dfrac{2}{x^2+14x+48}\)

\(A=\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{x+8}{x\left(x+8\right)}-\dfrac{x}{\left(x+8\right)}=\dfrac{8}{x\left(x+8\right)}\)

\(B=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{32}{1-x^{32}}\)

\(\dfrac{1}{x-1}-\dfrac{1}{x+1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{x+1-x+1}{x^2-1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{2}{x^2-1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{2\left(x^2+1-x^2+1\right)}{x^4-1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{4}{x^4-1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{4\left(x^4+1-x^4+1\right)}{x^8-1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{8}{x^8-1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{8\left(x^8+1-x^8+1\right)}{x^{16}-1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{16}{x^{16}-1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{16\left(x^{16}+1-x^{16}+1\right)}{x^{32}-1}\)

\(=\dfrac{32}{x^{32}-1}\)

1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(3x+9+4x-12=3x-7\)

\(\Leftrightarrow4x=-7+12-9=-4\)

hay \(x=-1\left(nhận\right)\)

2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)

\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(3x+12-4x+16=3x-4\)

\(\Leftrightarrow28-4x=-4\)

\(\Leftrightarrow4x=32\)

hay \(x=8\left(tm\right)\)

3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

Suy ra: \(5x^2-12+3x+3=5x^2-5x\)

\(\Leftrightarrow3x-9+5x=0\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(nhận\right)\)

\(A=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(A=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1+x\right)\left(1-x\right)}\right)+...+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)

Tiếp tục các bước như ở dòng 2 và 3 ta có :

\(A=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\dfrac{16\left(1-x^{16}\right)}{\left(1+x^{16}\right)\left(1-x^{16}\right)}\)

\(A=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}\)

\(A=\dfrac{32}{1-x^{32}}\)

1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)

Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)

\(\Leftrightarrow-3x-12-3+5x-x+4=0\)

\(\Leftrightarrow x=11\left(nhận\right)\)

2. ĐKXĐ: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)

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

\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)

\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)

Vậy pt vô nghiệm

3. ĐKXĐ: $x\neq \pm \frac{3}{2}$

PT \(\Leftrightarrow \frac{(x-5)(2x+3)-x(2x-3)}{(2x-3)(2x+3)}=\frac{1-6x}{(2x-3)(2x+3)}\)

\(\Rightarrow (x-5)(2x+3)-x(2x-3)=1-6x\)

\(\Leftrightarrow 2x^2-7x-15-2x^2+3x+6x-1=0\)

\(\Leftrightarrow 2x-16=0\Leftrightarrow x=8\) (thỏa mãn)

a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

a/ \(\dfrac{3-x}{12}=\dfrac{2x+2}{8}\)

\(< =>\dfrac{2\left(3-x\right)}{24}=\dfrac{3\left(2x+2\right)}{24}\)

\(< =>6-2x-6x-6=0\)

\(< =>-8x=0\)

\(< =>x=0\)

Vậy tập nghiệm.....

b/ \(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

Tìm ĐKXĐ của pt là: \(x\ne\pm4\)  (làm tắt, bạn làm rõ ra nhé)

\(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

\(< =>\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x^2+12\right)}{\left(x+4\right)\left(x-4\right)}\)

\(< =>x^2+3x+4x+12+x^2-3x-4x+12-2x^2-24=0\)

\(< =>0x=0\)

=> x có vô số nghiệm

Vậy ....

a) `(3-x)/12=(2x+2)/8`

`<=> (3-x)/12 =(x+1)/4`

`<=> 3-x=3(x+1)`

`<=>3-x=3x+3`

`<=> x=0`

Vậy `S={0}`.

b) ĐK: `x \ne \pm 4`

`(x+3)/(x-4)+(x-3)/(x+4)=(2(x^2+12))/(x^2-16)`

`<=> (x+3)(x+4)+(x-3)(x-4)=2(x^2+12)`

`<=> x^2+7x+12+x^2-7x+12=2x^2+24`

`<=> 0x=0`

Vậy PT có nghiệm với mọi x thỏa mãn điều kiện.

\(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

⇔\(\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x^2+12\right)}{\left(x-4\right)\left(x+4\right)}\)

⇔\(\dfrac{x^2+4x+3x+12}{\left(x-4\right)\left(x+4\right)}+\dfrac{x^2-4x-3x+12}{\left(x-4\right)\left(x+4\right)}=\dfrac{2x^2+24}{\left(x-4\right)\left(x+4\right)}\)

⇔\(\dfrac{x^2+7x+12}{\left(x-4\right)\left(x+4\right)}+\dfrac{x^2-7x+12}{\left(x-4\right)\left(x+4\right)}=\dfrac{2x^2+24}{\left(x-4\right)\left(x+4\right)}\)

⇒ \(x^2+7x+12+x^2-7x+12=2x^2+24\)

⇔ \(2x^2+24=2x^2+24\)

⇔ \(2x^2-2x^2=24-24\)

⇔ x=0

\(\dfrac{x}{x+4}+\dfrac{4}{x-4}-\dfrac{32}{x^2-16}\)

\(=\dfrac{x\left(x-4\right)+4\left(x+4\right)-32}{\left(x+4\right).\left(x-4\right)}\)

\(=\dfrac{x^2-4x+4x+16-32}{\left(x+4\right).\left(x-4\right)}\)

\(=\dfrac{x^2-16}{x^2-16}\)

\(=1\)

Ta có: \(\dfrac{x}{x+4}+\dfrac{4}{x-4}-\dfrac{32}{x^2-16}\)

\(=\dfrac{x\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\dfrac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{32}{x^2-16}\)

\(=\dfrac{x^2-4x+4x+16-32}{\left(x-4\right)\left(x+4\right)}\)

\(=\dfrac{x^2-16}{\left(x-4\right)\left(x+4\right)}\)

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