A=\(\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}.\)
a)Rút gọn A. b)Tìm x để A<2
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K
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KN
29 tháng 7 2019
a) \(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x^2-6x}{\left(x+3\right)\left(x-3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{3x^2-13x}{x^2-9}\)
14 tháng 10 2020
\(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
a) ĐK : x ≠ ±3
\(=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{x-3}\)
b) Để A < 2
=> \(\frac{3x}{x-3}< 2\)
<=> \(\frac{3x}{x-3}-2< 0\)
<=> \(\frac{3x}{x-3}-\frac{2x-6}{x-3}< 0\)
<=> \(\frac{3x-2x+6}{x-3}< 0\)
<=> \(\frac{x+6}{x-3}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}x+6>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-6\\x< 3\end{cases}}\Leftrightarrow-6< x< 3\)
2. \(\hept{\begin{cases}x+6< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -6\\x>3\end{cases}}\)( loại )
Vậy -6 < x < 3
5 tháng 8 2019
Mk chỉ cần câu trả lời thôi nên đừng vào cmt bậy nha các bác ;)))
12 tháng 12 2018
a, ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)
b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)
c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)
\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)
H
12 tháng 12 2018
a, ĐỂ A xác định :
\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)
\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)
\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)
\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{x-4}{x+3}\)
b
11 tháng 12 2019
ĐKXĐ:\(x\ne-3;x\ne3\)
\(A=\frac{5}{x+3}-\frac{2}{3-x}-\frac{3x^2-2x-9}{x^2-9}\)
\(=\frac{5}{x+3}+\frac{2}{x-3}-\frac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{5\left(x-3\right)+2\left(x+3\right)-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\frac{3x}{x+3}\)
b
\(\left|x-2\right|=1\Rightarrow x-2=1\left(h\right)x-2=-1\Rightarrow x=3;x=1\)
Tại \(x=3\) thì \(A=-\frac{3\cdot3}{3+3}=-\frac{9}{6}=-\frac{3}{2}\)
Tại \(x=1\) thì \(A=-1\cdot\frac{3}{1+3}=-\frac{3}{4}\)
c
Để A nguyên thì \(\frac{3x}{x+3}\) nguyên
\(\Rightarrow3x⋮x+3\)
\(\Rightarrow3\left(x+3\right)-9⋮x+3\)
\(\Rightarrow9⋮x+3\)
\(\Rightarrow x+3\in\left\{1;3;9;-1;-3;-9\right\}\)
\(\Rightarrow x\in\left\{-2;0;6;-4;-6;-12\right\}\)
30 tháng 12 2019
a) ĐKXĐ: \(\hept{\begin{cases}x+3\ne0\\3-x\ne0\\x^2-9\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ne-3\\x\ne3\\x\ne\pm3\end{cases}}\)
Ta có: A = \(\frac{x+1}{x+3}-\frac{x-1}{3-x}+\frac{2x-2x^2}{x^2-9}\)
A = \(\frac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+3\right)\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}+\frac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{2x-6}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
A = \(\frac{2}{x+3}\)
b) Để A nhận giá trị dương <=> 2 \(⋮\)x + 3
<=> x + 3 \(\in\)Ư(2) = {1; 2}
Lập bảng:
| x + 3 | 1 | 2 |
| x | -2 | -1 |
Vậy ....
