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1: \(B=\dfrac{6x+x^2-3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x}{x-3}\)
\(P=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)
a) P xác định \(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x-2\ne0\\2-x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}}\)
Vậy P xác định khi \(x\ne-3;x\ne2\)
b) \(P=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\left(x\ne-3;x\ne2\right)\)
\(\Leftrightarrow P=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)
\(\Leftrightarrow P=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow P=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow P=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)
Vậy \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)
c) \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)
Để P\(=-\frac{3}{4}\Rightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)
\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow4x+3x=6+16\)
\(\Leftrightarrow7x=22\)
\(\Leftrightarrow x=\frac{22}{7}\left(tm\right)\)
Vậy \(=\frac{22}{7}\)thì \(P=\frac{-3}{4}\)
d) \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2\right)\)
\(\Leftrightarrow P=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P nhận giá trị nguyên thì \(1-\frac{2}{x-2}\)nhận giá trị nguyên
\(\Leftrightarrow\frac{2}{x-2}\)nhận giá trị nguyên (1)
\(x\inℤ\Rightarrow x-2\inℤ\)(2)
\(\left(1\right)\left(2\right)\Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Ta có bảng giá trị
| x-2 | -2 | -1 | 1 | 2 |
| x | 0 | 1 | 3 | 4 |
Vậy \(x\in\left\{0;-1;3;4\right\}\)
a: ĐKXĐ: x<>1; x<>2; x<>-2; x<>-1
\(P=\dfrac{2017x+2017-2016x+2016-2014x-2016}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2015x+2017}{x^2-4}\)
a) ĐKXĐ : x \(\ne-2;x\ne1;x\ne0\)
\(A=\left(\frac{x}{x+2}-\frac{4}{x^2+2x}\right):\left(\frac{x^2-2x+1}{x^2-x}\right)=\left(\frac{x}{x+2}-\frac{4}{x\left(x+2\right)}\right):\left(\frac{\left(x-1\right)^2}{x\left(x-1\right)}\right)\)
\(=\frac{x^2-4}{x\left(x+2\right)}:\frac{x-1}{x}=\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}.\frac{x}{x-1}=\frac{x-2}{x}.\frac{x}{x-1}=\frac{x-2}{x-1}\)
b) Để A > 1
=> \(\frac{x-2}{x-1}>1\)
=> \(\frac{x-2}{x-1}-1>0\Rightarrow\frac{-1}{x-1}>0\Rightarrow x-1< 0\Rightarrow x< 1\)
Vậy để A > 1 thì x < 1 và x \(\ne-2;x\ne1;x\ne0\)
c) Ta có \(A=\frac{x-2}{x-1}=\frac{x-1-1}{x-1}=1-\frac{1}{x-1}\)
Để A \(\inℤ\Rightarrow\frac{1}{x-1}\inℤ\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)\Rightarrow x-1\in\left\{1;-1\right\}\)
Khi x - 1 = 1 => x = 2( tm)
Khi x - 1 =-1 => x = 0 (loại)
Vậy x = 2 thì A nguyên
a: ĐKXĐ: x<>1; x<>-1
b: \(A=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
c: Để A nguyên thì x+1-2 chia hết cho x+1
=>\(x+1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{0;-2;-3\right\}\)
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(P=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{1}\)
\(=\dfrac{x-2x+4+x-2}{x-2}\)
\(=\dfrac{2}{x-2}\)
b) Để P nguyên thì \(2⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{3;1;4;0\right\}\)