Cho biểu thức
A = (\(\frac{x}{x+3}\)+\(\frac{x}{x-3}\)-\(\frac{2}{x^2-9}\)) x \(\frac{x+3}{2x-2}\)
a. Tìm x để A = 2
b. Tìm x để A <0
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
24 tháng 6 2017
a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:
\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)
\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)
b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)
=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)
c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)
d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6
MN
25 tháng 2 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
\(\Leftrightarrow A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2\left(x+2\right)}{x-3}\)
\(\Leftrightarrow A=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x+4}{x-3}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow\frac{x+4}{x-3}\inℤ\)
\(\Leftrightarrow1+\frac{7}{x-3}\inℤ\)
\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
c) Để \(A=\frac{3}{5}\)
\(\Leftrightarrow\frac{x+4}{x-3}=\frac{3}{5}\)
\(\Leftrightarrow5x+20=3x-9\)
\(\Leftrightarrow2x+29=0\)
\(\Leftrightarrow x=-\frac{29}{2}\)
d) Để \(A< 0\)
\(\Leftrightarrow\frac{x+4}{x-3}< 0\)
\(\Leftrightarrow1+\frac{7}{x-3}< 0\)
\(\Leftrightarrow\frac{-7}{x-3}< 1\)
\(\Leftrightarrow-7< x-3\)
\(\Leftrightarrow x>-4\)
e) Để \(A>0\)
\(\Leftrightarrow\frac{x+4}{x-3}>0\)
\(\Leftrightarrow1+\frac{7}{x-3}>0\)
\(\Leftrightarrow\frac{-7}{x-3}>1\)
\(\Leftrightarrow-7>x-3\)
\(\Leftrightarrow x< -4\)
18 tháng 1 2021
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)
b) Ta có : \(A=\frac{x+4}{x-3}=\frac{x-3+7}{x-3}=1+\frac{7}{x-3}\)
Để A đạt giá trị nguyên thì \(\frac{7}{x-3}\)đạt giá trị nguyên
=> 7 ⋮ x - 3
=> x - 3 ∈ Ư(7) = { ±1 ; ±7 }
| x-3 | 1 | -1 | 7 | -7 |
| x | 4 | 2 | 10 | -4 |
So với ĐKXĐ ta thấy x = 4 , x = 10 , x = -4 thỏa mãn
Vậy với x ∈ { ±4 ; 10 } thì A đạt giá trị nguyên
18 tháng 1 2021
(....) dùng để nhìn được chữ số ở phân số cuối cùng thôi, ko dùng để làm gì.
( ác ) là từ ( các )
(gia strij) là từ ( giá trị )
a, \(A=\left(\frac{x}{x+3}+\frac{x}{x-3}-\frac{2}{x^2-9}\right).\frac{x+3}{2x-2}\)
\(=\frac{x^2-3x+x^2+3x-2}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{2\left(x-1\right)}=\frac{2\left(x-1\right)\left(x+1\right)\left(x+3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+3\right)}=\frac{x+1}{x-3}\)
Ta có : A = 2 hay \(\frac{x+1}{x-3}=2\Rightarrow x+1=2x-6\Leftrightarrow-x=-7\Leftrightarrow x=7\)(tmđk )
b, \(A< 0\Rightarrow\frac{x+1}{x-3}< 0\)
TH1 : \(\hept{\begin{cases}x+1< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>3\end{cases}}}\)( vô lí )
TH2 : \(\hept{\begin{cases}x+1>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x< 3\end{cases}\Rightarrow-1< x< 3}}\)
Kết hợp với đk ta được -1 < x < 3 ; x khác 1