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c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
a) Ta có: \(P=\left(\dfrac{3}{x+1}+\dfrac{x-9}{x^2-1}+\dfrac{2}{1-x}\right):\dfrac{x-3}{x^2-1}\)
\(=\left(\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x-9}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x-3}{x^2-1}\)
\(=\dfrac{3x-3+x-9-2x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x-3}\)
\(=\dfrac{2x-14}{x-3}\)
b) Ta có: \(x^2-9=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
Thay x=-3 vào biểu thức \(P=\dfrac{2x-14}{x-3}\), ta được:
\(P=\dfrac{2\cdot\left(-3\right)-14}{-3-3}=\dfrac{-20}{-6}=\dfrac{10}{3}\)
Vậy: Khi \(x^2-9=0\) thì \(P=\dfrac{10}{3}\)
c) Để P nguyên thì \(2x-14⋮x-3\)
\(\Leftrightarrow2x-6-8⋮x-3\)
mà \(2x-6⋮x-3\)
nên \(-8⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(-8\right)\)
\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
\(\Leftrightarrow x\in\left\{4;2;5;1;7;-1;11;-5\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;7;11;-5\right\}\)
Vậy: Để P nguyên thì \(x\in\left\{4;2;5;7;11;-5\right\}\)
Điều kiện xác định là `{(x-3 ne 0),(x(x-3) ne 0):}`
`<=>{(x ne 3),(x ne 0):}`
`=>bb A`
ĐCXĐ: \(\left\{{}\begin{matrix}x\ne0\\x-3\ne0\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)
\(C=\dfrac{-\left(x+1\right)+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
\(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Rightarrow C=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2\left(1-x\right)-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{-\left(1-x\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=-2.\dfrac{-1}{1-2x}\)
\(\Rightarrow C=\dfrac{2}{1-2x}\)
\(a,\left(12x^2y^2-6xy^2\right):3xy+2y=6xy^2\left(2x-1\right):3xy+2y=2y\left(2x-1\right)+2y=4xy-2y+2y=4xy\)
\(b,\dfrac{4}{x+1} + \dfrac{8}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{4\left(x-1\right)+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{4}{x-1}\)
\(c,\dfrac{1 }{x+1}- \dfrac{1}{x-1} +\dfrac{ 2x}{x^2-1} \)
\(=\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x-1-x-1+2x}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2x-2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2}{x+1}\)
\(a,=4xy-2y+2y=4xy\\ b,\dfrac{4}{x+1}+\dfrac{8}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\\ c,\dfrac{1}{x+1}-\dfrac{1}{x-1}+\dfrac{2x}{x^2-1}=\dfrac{x-1-x-1+2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2x-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+1}\)
1: \(P=\left(\dfrac{2x}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3x}\right)\)
\(=\left(\dfrac{2x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x\cdot\left(x-3\right)}\right)\)
\(=\dfrac{2x-x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2\left(x-3\right)-x+1}{x\left(x-3\right)}\)
\(=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(x-3\right)}{2x-6-x+1}\)
\(=\dfrac{x}{x-5}\)