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TH1:Nếu |x-1/2|+|x-y|=0
thì x-1/2=0
=>x=0+1/2=1/2
*)Nếu x=1/2 thì 1/2-y=0 hay y=1/2-0=1/2
còn th2 thì để mk nghĩ đã nha
a)\(\left(2x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\) hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< \frac{3}{2}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)
c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)
Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)
P=(1-1/1+2).(1-1/1+2+3)....(1-1/1+2+3+4+...+2011)
=[1-1/(2+1).2:2].[1-1/(3+1).3:2].....[1-1/(2011+1).2011:2]
=(1-2/2.3).(1-2/3.4)...(1-2/2011.2012)
=4/2.3.10/3.4....4046130/2011.2012
=1.4/2.3 .2.5/3.4 ....2010.2013/2011.2012
=1.2....2010/2.3...2011 .4.5....2013/3.4....2012
=1/2011.2013/3
=671/2011
a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
=>\(3x-\frac{1}{2}=0;\frac{1}{2}y+\frac{3}{5}=0\left(\left|3x-\frac{1}{2}\right|;\left|\frac{1}{2}y+\frac{3}{5}\right|\ge0\right)\)
=>\(x=\frac{1}{6};y=\frac{-6}{5}\)
b)\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
Ta lại có:
\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\)
=>\(\frac{3}{2}x+\frac{1}{9}=0;\frac{1}{5}y-\frac{1}{2}=0\Rightarrow x=-\frac{2}{27};y=\frac{5}{2}\)
Ko biết đúng hay sai nên đăng lại bài cho mấy bạn làm thử -'_'-
Hazz ....
\(\left(3x-1\right)\left(\frac{2}{3}x+\frac{1}{5}\right)\le0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x-1\le0\\\frac{2}{3}x+\frac{1}{5}\le0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x\le1\\\frac{2}{3}x\le-\frac{1}{5}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x\le\frac{1}{3}\\x\le-\frac{3}{10}\end{array}\right.\)
\(\Rightarrow x\le\frac{1}{3}\left(tm\right)\)
Vậy để \(\left(3x-1\right)\left(\frac{2}{3}x+\frac{1}{5}\right)\le0\) thì \(x\le\frac{1}{3}\)