Giúp mình giải bài này với, và ghi rõ nó là dòng bnhieu luôn nhé
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ĐKXĐ:\(x\ne\pm1\)
\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}=\dfrac{14}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{14}{\left(x+1\right)\left(x-1\right)}=0\\ \Rightarrow x^2-2x+1-x^2-2x-1-14=0\\ \Leftrightarrow-4x-14=0\\ \Leftrightarrow x=-\dfrac{7}{2}\left(tm\right)\)
\(ĐK:x\ne0;2\)
\(\Leftrightarrow\dfrac{x+2}{x}-\dfrac{2x+3}{2\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)\left(x+2\right)-x\left(2x+3\right)}{2x\left(x-2\right)}=0\)
\(\Leftrightarrow2\left(x^2-4\right)-x\left(2x+3\right)=0\)
\(\Leftrightarrow2x^2-8-2x^2-3x=0\)
\(\Leftrightarrow-3x=8\Leftrightarrow x=-\dfrac{8}{3}\left(tm\right)\)
\(\left|x+15\right|-2021=0\\ \Rightarrow\left|x+15\right|=2021\\ \Rightarrow\left[{}\begin{matrix}x+15=2021\\x+15=-2021\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2006\\x=-2036\end{matrix}\right.\)
\(\left|x+15\right|-2021=0\)
\(\Leftrightarrow\left|x+15\right|=2021\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=2021\\x+15=-2021\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2021-15\\x=-2021-15\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2006\\x=-2036\end{matrix}\right.\)
\(\left|x-2\right|+7=2x\\ \Leftrightarrow\left|x-2\right|=2x-7\left(x\ge\dfrac{7}{2}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-7\\x-2=7-2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
\(ĐK:x\ne\pm3\)
\(\Rightarrow\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=-\dfrac{3x-1}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)-\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=3x-1\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9+3x-1=0\)
\(\Leftrightarrow15x-1=0\)
\(\Leftrightarrow15x=1\)
\(\Leftrightarrow x=\dfrac{1}{15}\)
\(\dfrac{-3x-18+5x-10}{-15}-2\ge0\)
\(\Leftrightarrow\dfrac{2x-28+30}{-15}\ge0\Leftrightarrow\dfrac{2x+2}{-15}\ge0\)
\(\left\{{}\begin{matrix}2x+2< 0\\2x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x\ne-1\end{matrix}\right.\)
\(\Leftrightarrow3\left(2x-1\right)-2\left(3-x\right)=-1\)
=>6x-3-6+2x=-1
=>8x-9=-1
=>8x=8
hay x=1
1. F
Dẫn chứng: Vietnam's New Year is celebrated according to the Lunar calendar. It is especially known as Tet Nguyen Dan, or Tet.
2. T
Dẫn chứng: It begins between January twenty-first and February nineteen.
3. F
Dẫn chứng: Vietnamese people usually make preparations for the holiday several weeks beforehand.
4. T
Dẫn chứng: On the New Year's Eve, people sit up to midnight to see the New Year in, then they put on new clothes and give one another the greetings of the season.
5. T
Dẫn chứng: The first three days are the most important. Vietnamese people believe that how people act during those days will influence the whole year.
6. T
Dẫn chứng: As a result, they make every effort to avoid arguments and smile as much as possible.
1. T
2. T
3. F
4. T
5. T
6. T