Tìm x biết:
\(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
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11 tháng 9 2016
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: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)
Y
20 tháng 7 2019
1. Tìm x, biết :
a. ( x - \(\frac{3}{4}\)) \(^2\)= 0
=> x - \(\frac{3}{4}\)= 0
=> x = 0 + \(\frac{3}{4}\)
=> x = \(\frac{3}{4}\)
b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)
=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)
=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)
=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)
=> x = \(\frac{-1}{8}\)
c. \(\frac{\left(-2\right)^x}{16}=-8\)
=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)
=> ( -2)\(^x\)= -128
=> ( -2 ) \(^x\)= ( -2) \(^7\)
=> x = 7
14 tháng 10 2019
\(9,5-\frac{3}{4}\left|X-\frac{1}{3}\right|=6\frac{1}{3}-\frac{1}{3}\left|\frac{1}{3}-X\right|\)
\(\frac{19}{2}-\frac{3}{4}\left|X-\frac{1}{3}\right|=\frac{19}{3}-\frac{1}{3}\left|X-\frac{1}{3}\right|\)
\(\frac{19}{2}-\frac{3}{4}\left|X-\frac{1}{3}\right|+\frac{1}{3}\left|X-\frac{1}{3}\right|=\frac{19}{3}\)
\(\frac{19}{2}-\left(\frac{3}{4}\left|X-\frac{1}{3}\right|-\frac{1}{3}\left|X-\frac{1}{3}\right|=\frac{19}{3}\right)\)
\(\left|X-\frac{1}{3}\right|\left(\frac{3}{4}-\frac{1}{3}\right)=\frac{19}{2}-\frac{19}{3}\)
\(\frac{5}{12}\left|X-\frac{1}{3}\right|=\frac{19}{6}\)
\(\left|X-\frac{1}{3}\right|=\frac{19}{6}\div\frac{5}{12}\)
\(\left|X-\frac{1}{3}\right|=\frac{38}{5}\)
\(\Rightarrow\orbr{\begin{cases}X-\frac{1}{3}=\frac{38}{5}\\X-\frac{1}{3}=\frac{-38}{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{119}{15}\\x=\frac{-109}{15}\end{cases}}\)
Vậy.....................
P/s: sai thì bỏ qua nha!
NH
2
21 tháng 7 2015
Vì \(\left|x+\frac{3}{4}\right|\ge0;\left|y-\frac{1}{5}\right|\ge0;\left|x+y+z\right|\ge0\) với mọi x; y , z
nên để \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
thì \(\left|x+\frac{3}{4}\right|=\left|y-\frac{1}{5}\right|=\left|x+y+z\right|=0\)
=> \(x+\frac{3}{4}=0;y-\frac{1}{5}=0;x+y+z=0\)
+) x + 3/4 = 0 => x = -3/4
+) y - 1/5 = 0 => y =1/5
+) x + y + z = 0 => z = - x - y = 3/4 - 1/5 = 11/20
TD
0
25 tháng 10 2017
\(\left|x-\frac{5}{4}\right|-\left|x+\frac{2}{3}\right|=0\)
\(\left|x-\frac{5}{4}\right|=\left|x+\frac{2}{3}\right|\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{5}{4}=x+\frac{2}{3}\\x-\frac{5}{4}=-\frac{2}{3}-x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}0x=\frac{23}{12}\\2x=\frac{7}{12}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\text{không có giá trị x nào thỏa mãn}\\x=\frac{7}{24}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
=>\(\left(x-\frac{1}{3}\right)^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)
=>\(x-\frac{1}{3}=\frac{1}{2}\)hoặc \(x-\frac{1}{3}=-\frac{1}{2}\)
=>x=1/2+1/3hoawcj x=-1/2+1/3
=>x=5/6 hoặc x=-1/6