Câu hỏi của Vũ Thảo Anh

Question

giúp mk làm nốt mấy câu này với,cảm ơn trước nhá

ILoveMath · Accepted Answer

a) $D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)$$=\left(\dfrac{2}{x+2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)$$=\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)^2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\right)$$=\dfrac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{2x+4-4}{\left(x+2\right)^2}:\dfrac{-x}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{2x}{\left(x+2\right)^2}.\dfrac{\left(x-2\right)\left(x+2\right)}{-x}$$=\dfrac{-2.\left(x-2\right)}{x+2}$$x^2-5x+6=0\
\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\
\Rightarrow\left(x-2\right)\left(x-3\right)=0\
\Rightarrow\left[{}\begin{matrix}x-2=0\x-3=0\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=2\x=3\end{matrix}\right.$$P=\dfrac{-2.\left(x-2\right)}{x+2}$Thay $x=2$, ta có:$P=\dfrac{-2.\left(2-2\right)}{2+2}$    $=0$Thay $x=3$, ta có:$P=\dfrac{-2.\left(3-2\right)}{3+2}$    $=-\dfrac{2}{5}$ Câu trả lời: a) $D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)$$=\left(\dfrac{2}{x+2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)$$=\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)^2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\right)$$=\dfrac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{2x+4-4}{\left(x+2\right)^2}:\dfrac{-x}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{2x}{\left(x+2\right)^2}.\dfrac{\left(x-2\right)\left(x+2\right)}{-x}$$=\dfrac{-2.\left(x-2\right)}{x+2}$$x^2-5x+6=0\
\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\
\Rightarrow\left(x-2\right)\left(x-3\right)=0\
\Rightarrow\left[{}\begin{matrix}x-2=0\x-3=0\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=2\x=3\end{matrix}\right.$$P=\dfrac{-2.\left(x-2\right)}{x+2}$Thay $x=2$, ta có:$P=\dfrac{-2.\left(2-2\right)}{2+2}$    $=0$Thay $x=3$, ta có:$P=\dfrac{-2.\left(3-2\right)}{3+2}$    $=-\dfrac{2}{5}$

ILoveMath · Answer

D nguyên âm $\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\left(x-2\right)< 0\x+2>0\end{matrix}\right.\\left\{{}\begin{matrix}-2\left(x-2\right)>0\x+2< 0\end{matrix}\right.\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\x>-2\end{matrix}\right.\\left\{{}\begin{matrix}x< 2\x< -2\end{matrix}\right.\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x>2\x< -2\end{matrix}\right.$ Câu trả lời: D nguyên âm $\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\left(x-2\right)< 0\x+2>0\end{matrix}\right.\\left\{{}\begin{matrix}-2\left(x-2\right)>0\x+2< 0\end{matrix}\right.\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\x>-2\end{matrix}\right.\\left\{{}\begin{matrix}x< 2\x< -2\end{matrix}\right.\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x>2\x< -2\end{matrix}\right.$

Nguyễn Lê Phước Thịnh · Answer

a:Ta có: $D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)$$=\dfrac{2x+4-4}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{2x}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-x}$$=\dfrac{-\left(x-2\right)}{x+2}$b: Ta có: $x^2-5x+6=0$$\Leftrightarrow\left(x-2\right)\left(x-3\right)=0$hay x=3Thay x=3 vào D, ta được:$D=\dfrac{-\left(3-2\right)}{3+2}=-\dfrac{1}{5}$Câu trả lời: a:Ta có: $D=\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)$$=\dfrac{2x+4-4}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{2x}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)\left(x-2\right)}{-x}$$=\dfrac{-\left(x-2\right)}{x+2}$b: Ta có: $x^2-5x+6=0$$\Leftrightarrow\left(x-2\right)\left(x-3\right)=0$hay x=3Thay x=3 vào D, ta được:$D=\dfrac{-\left(3-2\right)}{3+2}=-\dfrac{1}{5}$

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