Violympic toán 8

NH

Cho bt: A = \((\frac{2-x}{x+3}-\frac{3-x}{x+2}+\frac{2-x}{x^2+5x+6}):(1-\frac{x}{x-1})\)

a, rút gọn A

b, Tìm x để A = 0 ; A > 0

TP
24 tháng 6 2019 lúc 15:41

a) \(A=\left(\frac{2-x}{x+3}-\frac{3-x}{x+2}+\frac{2-x}{x^2+5x+6}\right):\left(1-\frac{x}{x-1}\right)\)

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

\(A=\frac{4-x^2-9+x^2+2-x}{\left(x+3\right)\left(x+2\right)}\cdot\frac{1-x}{1}\)

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

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

b) \(A=0\Leftrightarrow\frac{x-1}{x+2}=0\Leftrightarrow x=1\)( không thỏa mãn ĐKXĐ )

\(A>0\Leftrightarrow\frac{x-1}{x+2}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

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H24
24 tháng 6 2019 lúc 16:25

\(A=\left(\frac{2-x}{x+3}-\frac{3-x}{x+2}+\frac{2}{x^2+5x+6}\right):\left(1-\frac{x}{x-1}\right)=\left(\frac{\left(2-x\right)\left(2+x\right)}{x^2+5x+6}-\frac{\left(3-x\right)\left(3+x\right)}{x^2+5x+6}+\frac{2}{x^2+5x+6}\right):\left(\frac{-1}{x-1}\right)=\left(\frac{4-x^2}{x^2+5x+6}-\frac{9-x^2}{x^2+5x+6}+\frac{2}{x^2+5x+6}\right):\left(\frac{-1}{x-1}\right)=\frac{-3}{x^2+5x+6}:\frac{-1}{x-1}=\frac{x-1}{x^2+5x+6}\)

\(A=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(A>0\)

\(TH1:+,\left\{{}\begin{matrix}x-1< 0\\x^2+5x+6< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 1\\\left(x+2,5\right)^2< 0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\-0,5< x+2,5< 0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\-3< x< -2\end{matrix}\right.\Rightarrow-3< x< -2\)

\(+,\left\{{}\begin{matrix}x-1>0\\x^2+5x+6>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>1\\\left(x+2,5\right)^2>0,25\end{matrix}\right.\)

\(\left(x+2,5\right)>0,25\Leftrightarrow\left[{}\begin{matrix}x+2,5>0,5\\x+2,5< -0,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-2\\x< -3\end{matrix}\right.\Leftrightarrow x>1\)

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H24
24 tháng 6 2019 lúc 16:28

baimk bi sai r eo

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