\(\dfrac{6}{x^2-3x}\)  = \(\dfrac{A}{x}\) + \(\dfrac{B}{x-3}\) (nếu đúng với mọi \(x\) ≠0; 3 thì làm như sau)

Đkxđ: \(\left\{{}\begin{matrix}x^2-3x\ne0\\x\ne0\\x-3\ne0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x\left(x-3\right)\ne0\\x\ne0\\x\ne3\end{matrix}\right.\)  ⇒ \(\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)

\(\dfrac{A}{x}\) + \(\dfrac{B}{x-3}\) = \(\dfrac{A.\left(x-3\right)}{x.\left(x-3\right)}\) + \(\dfrac{B.x}{x\left(x-3\right)}\) = \(\dfrac{Ax-3A+Bx}{x\left(x-3\right)}\)

⇒\(\dfrac{6}{x^2-3x}\) = \(\dfrac{6}{x.\left(x-3\right)}\) = \(\dfrac{Ax-3A+Bx}{x.\left(x-3\right)}\)

⇒  \(\dfrac{6}{x\left(x-3\right)}\) - \(\dfrac{Ax-3A+Bx}{x\left(x-3\right)}\) = 0

        \(\dfrac{1}{x\left(x-3\right)}\).[6 -  (A\(x\) - 3A + B\(x\))] = 0 

                  ⇒   6 - A\(x\) + 3A - B\(x\) = 0 

                 ⇒ - \(x\).( A + B) + 6 + 3A = 0 (1)

        (1) đúng với ∀ \(x\) ≠0; 3 ⇔ \(\left\{{}\begin{matrix}A+B=0\\6+3A=0\end{matrix}\right.\)

                                            ⇔ \(\left\{{}\begin{matrix}A=-B\\3A=-6\end{matrix}\right.\)

                                            ⇔ \(\left\{{}\begin{matrix}A=-B\\A=-6:3\end{matrix}\right.\)

                                           ⇔ \(\left\{{}\begin{matrix}B=-A\\A=-2\end{matrix}\right.\)

                                             ⇒ \(\left\{{}\begin{matrix}B=2\\A=-2\end{matrix}\right.\)

Vậy A = -2; B = 2

\(\dfrac{6}{x^2-3x}=\dfrac{A}{x}+\dfrac{B}{x-3}\left(x\ne0;x\ne3\right)\)

\(\Leftrightarrow\dfrac{6}{x\left(x-3\right)}=\dfrac{A\left(x-3\right)}{x\left(x-3\right)}+\dfrac{Bx}{x\left(x-3\right)}\)

\(\Leftrightarrow6=A\left(x-3\right)+Bx\)

\(\Leftrightarrow6=Ax-3A+Bx\) 

\(\Leftrightarrow0x+6=\left(A+B\right)x-3A\)

\(\Leftrightarrow\left\{{}\begin{matrix}A+B=0\\-3A=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=-B\\A=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}B=2\\A=-2\end{matrix}\right.\)

Bài 3:

a, rút gọn P = \(\dfrac{x^2}{x+1}\) + \(\dfrac{2.\left(x-1\right)}{x}\) + \(\dfrac{x+2}{x^2+x}\) với \(x\ne0;x\ne-1\)

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

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

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

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

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

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

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

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

P = \(x\) + 1

b, Thay \(x\) = 1 vào biểu thức P = \(x\) + 1 ta có:

P = 1 + 1

P = 2

a) \(A+\dfrac{1}{x+1}=\dfrac{3x+1}{x^2-2x+1}-\dfrac{x+3}{x^2-1}\left(x\ne\pm1\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(P=\dfrac{2}{x^2+x+1}\)

a, \(\dfrac{x+3}{3x-3}\) + \(\dfrac{2-x}{4x-4}\) 

đkxđ: \(\left\{{}\begin{matrix}3x-3\ne0\\4x-4\ne0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}3.\left(x-1\right)\ne0\\4.\left(x-1\right)\ne0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-1\ne0\\x-1\ne0\end{matrix}\right.\) ⇒ \(x\) ≠1

A = \(\dfrac{x+3}{3x-3}\) + \(\dfrac{2-x}{4x-4}\)

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

A = \(\dfrac{4.\left(x+3\right)}{4.3.\left(x-1\right)}\) + \(\dfrac{3.\left(2-x\right)}{3.4.\left(x-1\right)}\)

A = \(\dfrac{4x+12}{12\left(x-1\right)}\) + \(\dfrac{6-3x}{12\left(x-1\right)}\)

A = \(\dfrac{4x+12+6-3x}{12\left(x-1\right)}\)

A = \(\dfrac{\left(4x-3x\right)+\left(12+6\right)}{12\left(x-1\right)}\)

A = \(\dfrac{x+18}{12.\left(x-1\right)}\)

b, \(\dfrac{3}{x+2}-\dfrac{6}{x-1}\) 

Đkxđ: \(\left\{{}\begin{matrix}x+2\ne0\\x-1\ne0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x\ne-2\\x\ne1\end{matrix}\right.\)

B = \(\dfrac{3}{x+2}\) - \(\dfrac{6}{x-1}\)

B = \(\dfrac{3.\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}\) - \(\dfrac{6.\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

B = \(\dfrac{3x-3}{\left(x+2\right)\left(x-1\right)}\) - \(\dfrac{6x+12}{\left(x-1\right)\left(x+2\right)}\)

B = \(\dfrac{3x-3-\left(6x+12\right)}{\left(x+2\right)\left(x-1\right)}\)

B = \(\dfrac{3x-3-6x-12}{\left(x+2\right)\left(x-1\right)}\)

B = \(\dfrac{\left(3x-6x\right)-\left(12+3\right)}{\left(x+2\right)\left(x-1\right)}\)

B = \(\dfrac{-3x-15}{\left(x+2\right)\left(x-1\right)}\)

and

I have to tidy my bedroom because I have to dust the living room.

   2\(x^2\)y² - 6\(\sqrt{2}\)\(xy\) + 9

= (\(\sqrt{2}\).\(x.y\))² - 2.\(\sqrt{2}\)\(xy\).3 + 3²

= (\(\sqrt{2}\)\(xy\)  - 3)²

\(M_{H_2}=1.2=2g/mol\\ M_{O_2}=16.2=32g/mol\\ M_{CO_2}=12+16.2=44g/mol\\ M_{Al_2O_3}=27.2+16.3=102g/mol\\ M_{Fe}=56g/mol\\ M_{Mg}=24g/mol\)

a) PTHH \(S+O_2\underrightarrow{t^o}SO_2\)

b) Khí SO₂ chính là chất làm cho chuột chết. Hợp chất này tên là lưu huỳnh đi-ô-xít (tên Tiếng Anh là sulfur dioxide). Đây là một oxit axit.