Tìm x biết
\(\frac{3}{5}-\left|x+1\right|=\frac{1}{2}\)
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1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}=\frac{2.15}{5.11}=\frac{6}{11}\)
Vậy x = 6/11
a) \(\frac{1}{3}.x+\frac{2}{5}.\left(x-1\right)=0\)
\(\frac{1}{3}.x+\frac{2}{5}.x-\frac{2}{5}=0\)
\(x.\left(\frac{1}{3}+\frac{2}{5}\right)-\frac{2}{5}=0\)
\(x.\frac{11}{15}-\frac{2}{5}=0\)
\(x.\frac{11}{15}=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{11}{15}\)
\(x=\frac{6}{11}\)
b) \(3.\left(x-\frac{1}{2}\right)-5.\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)
\(3x-5x-\left(\frac{3}{2}+3\right)=x+\frac{1}{5}\)
\(-2x-\frac{9}{2}=x+\frac{1}{5}\)
\(\Rightarrow-2x-x=\frac{1}{5}+\frac{9}{2}\)
\(-3x=\frac{47}{10}\)
\(x=\frac{47}{10}:\left(-3\right)\)
\(x=\frac{-47}{30}\)
\(3\left(2x-\frac{5}{4}\right)=\left(3-1\frac{1}{2}\right)\left(x-\frac{1}{2}\right)\)
\(\Leftrightarrow6x-\frac{15}{4}=\frac{3}{2}x+\frac{1}{12}\)
\(\Leftrightarrow\frac{9}{2}x+\frac{3}{4}=\frac{15}{4}\)
\(\Leftrightarrow\frac{9}{2}x=3\)
\(\Leftrightarrow x=\frac{2}{3}\)
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35
Bài 1:
a) (2x-3). (x+1) < 0
=>2x-3 và x+1 ngược dấu
Mà 2x-3<x+1 với mọi x
\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)
b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu
Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)
Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
=>....
Bài 2:
\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\cdot\frac{998}{3003}\)
\(=\frac{499}{3003}\)
\(\left(\frac{1}{4}x-1\right)-\left(\frac{2}{3}x-1\right)+\left(\frac{4}{5}x-1\right)=\frac{2}{3}\)
\(\frac{1}{4}x-1-\frac{2}{3}x+1+\frac{4}{5}x-1\)\(=\frac{2}{3}\)
\(\left(\frac{1}{4}x-\frac{2}{3}x+\frac{4}{5}x\right)+1-1-1\)\(=\frac{2}{3}\)
\(\frac{23}{60}x-1\)\(=\frac{2}{3}\)
\(\frac{23}{60}x=\frac{2}{3}+1\)
\(\frac{23}{60}x=\frac{5}{3}\)
\(x=\frac{5}{3}:\frac{23}{60}=\frac{100}{23}\)
Vậy x=\(\frac{100}{23}\)
=> \(\frac{1}{7}x-\frac{2}{7}=0\)
=> \(x=2\)
hoặc
\(-\frac{1}{5}x+\frac{3}{5}=0\)
=> x = 3
hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)
=> x = -4
Vậy x = -4; 2; 3
\(\frac{3}{5}-\left|x+1\right|=\frac{1}{2}\)
\(\left|x+1\right|=\frac{3}{5}-\frac{1}{2}\)
\(\left|x+1\right|=\frac{1}{10}\)
\(\Rightarrow\)\(\hept{\begin{cases}x+1=\frac{1}{10}\\x+1=\frac{-1}{10}\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=\frac{1}{10}-1\\x=\frac{-1}{10}-1\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=-\frac{9}{10}\\x=\frac{-11}{10}\end{cases}}\)
Vậy \(x\in\left\{-\frac{9}{10};-\frac{11}{10}\right\}\)
\(|x+1|=\frac{3}{5}-\frac{1}{2}\)
\(|x+1|=\frac{1}{10}\)
Th1: \(x+1=\frac{1}{10}\)
\(x=\frac{1}{10}-1\)
\(x=-\frac{9}{10}\)
Th2: \(x+1=\frac{-1}{10}\)
\(x=-\frac{1}{10}-1\)
\(x=-\frac{11}{10}\)
Vậy \(x=-\frac{9}{10};-\frac{11}{10}\)