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1, a/ \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy .............
b/ \(\left|x\right|=3,12\Leftrightarrow\left[{}\begin{matrix}x=3,12\\x=-3,12\end{matrix}\right.\)
Vậy ...........
c/ \(\left|x\right|=0\Leftrightarrow x=0\)
Vậy ..........
d/ \(\left|x\right|=2\dfrac{1}{7}\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\dfrac{1}{7}\\x=-2\dfrac{1}{7}\end{matrix}\right.\)
Vậy ..............
2, a/ \(\left|x\right|=2,1\Leftrightarrow\left[{}\begin{matrix}x=2,1\\x=-2,1\end{matrix}\right.\)
Vậy ...........
b/ \(\left|x\right|=\dfrac{17}{9}\) ; \(x< 0\)
\(\Leftrightarrow x=-\dfrac{17}{9}\)
Vậy ..........
c/ \(\left|x\right|=1\dfrac{2}{5}\Leftrightarrow\left[{}\begin{matrix}x=1\dfrac{2}{5}\\x=-1\dfrac{2}{5}\end{matrix}\right.\)
Vậy ...........
d/ \(\left|x\right|=0,35\) ; \(x>0\Leftrightarrow x=0,35\)
3, a/ \(\left|x-1,7\right|=2,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy ...........
b/ \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy ...........
\(\left|x\right|=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}\)hoặc \(-\frac{1}{2}\)
\(\left|x\right|=3,12\)
\(\Rightarrow x=3,12\)hoặc \(-3,12\)
\(\left|x\right|=0\)
\(\Rightarrow x=0\)
\(\left|x\right|=\frac{2}{\frac{1}{7}}=14\)
\(\Rightarrow x=14\)hoặc \(-14\)
\(\left|x\right|=\frac{1}{2}\)
\(\Rightarrow x=\pm\frac{1}{2}\)
\(\left|x\right|=3,12\)
\(\Rightarrow x=\pm3,12\)
\(\left|x\right|=0\)
\(\Rightarrow x=0\)
\(\left|x\right|=2\frac{1}{7}\)
\(\Rightarrow x=\pm2\frac{1}{7}\)
/-3/ = /3/
/1,3/ > /0,5/
/-100/ > /20/
bài 2: tìm x
/x/ = \(\frac{1}{2}\)
\(x=\orbr{\begin{cases}\frac{1}{2}\\\frac{-1}{2}\end{cases}}\)
/x/ = 3,12
\(x=\orbr{\begin{cases}3,12\\-3,12\end{cases}}\)
/x/ = 0
=> x = 0
/x/ = \(2\frac{1}{7}\)
=> /x/ = \(\frac{15}{7}\)
=> \(x=\orbr{\begin{cases}\frac{15}{7}\\\frac{-15}{7}\end{cases}}\)
chúc bn học tốt
a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
| x - 2 | 1 | -1 | 7 | -7 |
| x | 3 | 1 | 9 | -5 |
Vậy....
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
3) a) \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=\dfrac{1}{2};x=\dfrac{-1}{2}\)
b) \(\left|x\right|=3,12\Leftrightarrow\left[{}\begin{matrix}x=3,12\\x=-3,12\end{matrix}\right.\) vậy \(x=3,12;x=-3,12\)
c) \(\left|x\right|=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-0\end{matrix}\right.\) \(\Rightarrow x=0\) vậy \(x=0\)
d) th1: \(\left|x\right|=2\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
th2: \(\left|x\right|=\dfrac{1}{7}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=\dfrac{-1}{7}\end{matrix}\right.\)
vậy \(x=2;x=-2;x=\dfrac{1}{7};x=\dfrac{-1}{7}\)
cảm ơn nhé