\(\frac{1}{2}=|2x-1|\)
giúp mk nha
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\(\frac{-6}{3}\left[x-\frac{1}{4}\right]=2x-1\)
\(-2x-\left[\frac{1}{4}.-2\right]=2x-1\)\
\(-2x-\frac{-1}{2}=2x-1\)
\(2x--2x=1-\frac{-1}{2}\)
\(\)\(4x=\frac{3}{2}\)
\(x=\frac{3}{2}:4\)
\(x=\frac{3}{8}\)
<=>\(\sqrt{2x+1}-1>\frac{2x}{2x}+\frac{2x}{2}\)=1+x
=>x+1-\(\sqrt{2x+1}\)+1<0
nhân 2 vô 2x+1-\(2\sqrt{2x+1}\)+1+1<0
Ta có\(-\frac{2}{3}\) \(X\) (\(X\) \(-\frac{1}{4}\) ) = \(\frac{1}{3}\)\(X\) (\(2X-1\) )
\(\Rightarrow\) \(\frac{-2}{3}\) \(X^2\)\(+\) \(\frac{1}{6}\) \(X\) = \(\frac{2}{3}\) \(X^2\) \(-\) \(\frac{1}{3}\) \(X\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{6}\) = \(\frac{2}{3}\) \(X\) \(-\) \(\frac{1}{3}\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{6}\) \(+\) \(\frac{1}{3}\) = \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{2}\) = \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{1}{2}\) = \(\frac{2}{3}\) \(X\) \(+\) \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{1}{2}\) = \(X\) (\(\frac{2}{3}\) \(+\) \(\frac{2}{3}\) )
\(\Rightarrow\) \(\frac{1}{2}\) = \(\frac{4}{3}\) \(X\)
\(\Rightarrow\) \(X\) = \(\frac{1}{2}\) \(\div\) \(\frac{4}{3}\)
\(\Rightarrow\) \(X\) = \(\frac{3}{8}\)
Có gì không hiểu cứ hỏi tớ nhá !
a ) MTC : \(2x\left(x+3\right)\left(x-3\right)\)
\(\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{3-2x}{x^2-9}=\frac{3-2x}{\left(x-3\right)\left(x+3\right)}=\frac{2x\left(3-2x\right)}{2x\left(x+3\right)\left(x-3\right)}\)
b ) MTC : \(2\left(-x\right)\left(x-1\right)^2\)
\(\frac{2x-1}{x-x^2}=\frac{2x-1}{-x\left(x-1\right)}=\frac{2\left(2x-1\right)\left(x-1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
\(\frac{x+1}{2-4x+2x^2}=\frac{x+1}{2\left(x^2-2x+1\right)}=\frac{-x\left(x+1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
\(\frac{2x}{3}-\frac{1}{6}.x-\frac{x+4}{2}=0\)
\(\Rightarrow\frac{4x}{6}-\frac{x}{6}-\frac{3x+12}{6}=0\)
\(\Rightarrow\frac{4x-x-3x-12}{6}=0\)
\(\Rightarrow0-12=0\Rightarrow-12=0\left(\text{vô lí}\right)\)
Vậy không có x thỏa mãn
\(\left|2x-1\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}2x-1=\frac{1}{2}\\2x-1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{4}\end{cases}}}\)
Vậy x = .................
Tham khảo
|2x - 1| = \(\frac{1}{2}\)
<=> \(\orbr{\begin{cases}2x-1=\frac{1}{2}\\2x-1=-\frac{1}{2}\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{4}\end{cases}}\)
Vậy ...