\(\frac{a}{x-2}+\frac{b}{\left(x+1\right)^2}=\frac{a\left(x+1\right)^2+b\left(x-2\right)}{\left(x-2\right)\left(x+1\right)^2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)

\(\Rightarrow\frac{x^2+5}{x^3-3x-2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)

Đồng nhất hệ số, ta có :

\(\hept{\begin{cases}a=1\\2a+b=0\\a-2b=5\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=-2\end{cases}}}\)

cái thứ 2 tương tự

=.= tương tự là seo????

ĐKXĐ : \(x\ne\left\{-1;2\right\}\)

\(\frac{x^2+5}{x^3-3x-2}=\frac{a}{x-2}+\frac{b}{\left(x+1\right)^2}\)

\(\Leftrightarrow\frac{x^2+5}{\left(x-2\right)\left(x+1\right)^2}=\frac{a\left(x+1\right)^2+b\left(x-2\right)}{\left(x-2\right)\left(x+1\right)^2}\)

\(\Leftrightarrow\frac{x^2+5}{\left(x-2\right)\left(x+1\right)^2}=\frac{ax^2+2ax+a+bx-2b}{\left(x-2\right)\left(x+1\right)^2}\)

\(\Leftrightarrow\frac{x^2+5}{\left(x-2\right)\left(x+1\right)^2}=\frac{ax^2+x\left(2a+b\right)+\left(a-2b\right)}{\left(x-2\right)\left(x+1\right)^2}\)

Đồng nhất hệ số ta được : \(\hept{\begin{cases}a=1\\2a+b=0\\a-2b=5\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=-2\end{cases}}}\)

Vậy \(a=1;b=-2\)

giúp tớ vs bài này khó wa

câu 1

a)\(ĐKXĐ:x^3-8\ne0=>x\ne2\)

b)\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2-2x+4\right)}=\frac{3}{x-2}\left(#\right)\)

Thay \(x=\frac{4001}{2000}\)zô \(\left(#\right)\)ta được

\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)

c) Để phân thức trên có giá trị nguyên thì :

\(3⋮x-2\)

=>\(x-2\inƯ\left(3\right)=\left(\pm1\pm3\right)\)

=>\(x\in\left\{1,3,-1,5\right\}\)

zậy ....

câu 2)

a) \(8\left(3x-2\right)-14x=2\left(4-7x\right)+15x\)

=>\(24x-16-14x=8-14x+15x\)

=>\(24x-14x+14x-15x=8+16\)

=>\(9x=24=>x=\frac{24}{9}=\frac{8}{3}\)

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

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

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

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

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

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

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

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

A = \(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

A = \(\frac{x+1}{x-1}\) (Đk: \(x-1\ge0\) => x \(\ge\)1)

b) Ta có: A = \(\frac{x+1}{x-1}=\frac{\left(x-1\right)+2}{x-1}=1+\frac{2}{x-1}\)

Để A \(\in\)Z <=> 2 \(⋮\)x - 1

<=> x - 1 \(\in\)Ư(2) = {1; -1; 2; -2}

<=> x \(\in\){2; 0; 3; -1}

c) Ta có: A < 0

=> \(\frac{x+1}{x-1}< 0\)

=> \(\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\) hoặc \(\hept{\begin{cases}x+1>0\\x-1< 0\end{cases}}\)

=> \(\hept{\begin{cases}x< -1\\x>1\end{cases}}\)(loại) hoặc \(\hept{\begin{cases}x>-1\\x< 1\end{cases}}\) 

=> -1 < x < 1

Edogawa Conan

Thiếu dòng đầu  \(ĐKXĐ:\hept{\begin{cases}x\ne1\\x\ne-2\\x\ne0\end{cases}}\)

ĐKXĐ : \(\) x # +1 ; x # - 1 ; x # -2 ; x # 0 ; x # 2

 Ta có: \(A=\frac{3x^2+3x-3}{x^2+x-2}-\frac{x+1}{x+2}+\frac{x-2}{x}.\left(\frac{1}{1-x}-1\right)\)

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

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

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

  \(=\frac{3x^2+3x-3}{x^2+x-2}-\frac{2x^2-5}{x^2+x-2}\)

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

   \(\frac{x+1}{x-1}\)

b. Ta có:  \(A=\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\)

Để A nhận giá trị nguyên thì: \(2⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(2\right)\)

  +) x - 1 = 1 => x = 2   (loại)

  +) x - 1 = 2 => x = 3  

  +) x - 1 = -1 => x = 0  (loại)

  +) x - 1 = -2 => x = -1    (loại)

Vậy x = 3 là giá trị cần tìm.

c.  \(A< 0\Leftrightarrow\frac{x+1}{x-1}< 0\)

\(\Leftrightarrow\hept{\begin{cases}x+1>0\\x-1< 0\end{cases}}\)   hoặc \(\hept{\begin{cases}x+1< 0\\x-1>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>-1\\x< 1\end{cases}}\)    hoặc   \(\hept{\begin{cases}x< -1\\x>1\end{cases}}\)(vô lý)

Vậy \(-1< x< 1\) và x # 0 là giá trị cần tìm