Giải:

a) \(8\left(3x-2\right)-13x=5\left(12-3x\right)+7x\)

\(\Leftrightarrow24x-16-13x=60-15x+7x\)

\(\Leftrightarrow24x-13x+15x-7x=60+16\)

\(\Leftrightarrow19x=76\)

\(\Leftrightarrow x=\dfrac{76}{19}=4\)

Vậy ...

b) \(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\) (1)

ĐKXĐ: \(x\ne\pm2\)

\(\left(1\right)\Leftrightarrow\dfrac{5x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow5x\left(x-2\right)-3\left(x+2\right)+3x^2+6=0\)

\(\Leftrightarrow5x^2-10x-3x-6+3x^2+6=0\)

\(\Leftrightarrow8x^2-13x=0\)

\(\Leftrightarrow x\left(8x-13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{13}{8}\left(TM\right)\end{matrix}\right.\)

Vậy ...

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

ĐKXĐ: \(x\ne-1;x\ne3\)

\(\left(2\right)\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)

\(\Leftrightarrow x\left(x+1+x-3\right)=4x\)

\(\Leftrightarrow x\left(2x-2\right)=4x\)

\(\Leftrightarrow2x-2=4\)

\(\Leftrightarrow x=3\)

Vậy ...

Sửa đề:

 \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

ĐKXĐ: \(x\notin\left\{1;3;8;20\right\}\)

PT=>\(-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

=>\(-\dfrac{1}{x-4}=-\dfrac{3}{4}\)

=>\(x-1=\dfrac{4}{3}\)

=>\(x=\dfrac{4}{3}+1=\dfrac{7}{3}\)(nhận)

a: \(=\dfrac{4xy+x^2-2xy+y^2}{2\left(x+y\right)\left(x-y\right)}\cdot\dfrac{2x}{x+y}-\dfrac{y}{x-y}\)

\(=\dfrac{x}{x-y}-\dfrac{y}{x-y}=1\)

b: \(=\dfrac{x^2+x-2-2x^2-2x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{3\left(x+1\right)}{x}+\dfrac{4x^2+x+7}{x\left(x-1\right)}\)

\(=\dfrac{-x^2-x-2}{\left(x-1\right)}\cdot\dfrac{3}{x}+\dfrac{4x^2+x+7}{x\left(x-1\right)}\)

\(=\dfrac{4x^2+x+7-3x^2-3x-6}{x\left(x-1\right)}=\dfrac{x^2-2x+1}{x\left(x-1\right)}=\dfrac{x-1}{x}\)

c: \(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}\)

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

giải hết đống này chắc @@ quá,để tối đi,giờ t đi làm mấy bài ngắn ngắn

a) x²+\(\dfrac{2x}{x-1}\)=8(ĐKXĐ : x ≠ 1

⇔ x²(x-1)+2x =8⇔ x³ - x² +2x - 8=0

⇔x³ - 2³ -x²+2x =0⇔ (x-2)(x² +x+1) -x(x-2)

⇔(x-2)(x² +1)⇒x =2

x² +1 =0⇒x² -1⇒x ∈∅(loại)

vậy x =2

Giải phương trình :

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

ĐKXĐ : \(x-1\ne0\Rightarrow x\ne1\)

Ta có : \(x^2+\dfrac{2x}{x-1}=8\)

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

\(\Rightarrow x^2\left(x-1\right)+2x=8\left(x-1\right)\)

\(\Leftrightarrow x^3-x^2+2x=8x-8\)

\(\Leftrightarrow x^3-x^2+2x-8x=-8\)

\(\Leftrightarrow x^3-x^2-6x+8=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(6x-8\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)-2\left(3x-4\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x-1\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x-1=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)

Đối chiếu với ĐKXĐ ta được \(x\in\left\{\sqrt{2};\dfrac{4}{3}\right\}\) thỏa mãn.

a) điều kiện xác định : \(x\ne0\)

ta có : \(A=\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow A=\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)

b) điều kiện : \(x\notin\left\{-4;-5;-6;-7\right\}\)

\(B=\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow B=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow B=\dfrac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow B=\dfrac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow54=x^2+11x+28\Leftrightarrow x^2+11x+28-54=0\)

\(\Leftrightarrow x^2+11x-26=0\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2;x=-13\)

a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)

( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)

(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)

1/x +1/x+4

2x+4/x(x+4)

Câu b bạn tách các mẫu thành nhân tử rồi làm như câu a nhé

b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)

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

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

\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)

\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)

c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)

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

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

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