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Ix-1,7I = 2,3
TH1: x - 1,7 = 2,3
=> x = 2,3 + 1,7
=> x = 4
TH2 : x - 1,7 = -2,3
=> x = -2,3 + 1,7
=> x = -0,6
b) Ix + 3/4I - 1/3 = 0
=> Ix + 3/4I = 0 + 1/3
=> x + 3/4 = 1/3
=> x = 1/3 - 3/4
=> x = -5/12
a.
\(\left|x-1,7\right|=2,3\)
\(x-1,7=\pm2,3\)
TH1:
\(x-1,7=2,3\)
\(x=2,3+1,7\)
\(x=4\)
TH2:
\(x-1,7=-2,3\)
\(x=-2,3+1,7\)
\(x=-0,6\)
Vậy x = 4 hoặc x = -0,6
b.
\(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)
\(\left|x+\frac{3}{4}\right|=\frac{1}{3}\)
\(x+\frac{3}{4}=\pm\frac{1}{3}\)
TH1:
\(x+\frac{3}{4}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{3}{4}\)
\(x=\frac{4-9}{12}\)
\(x=-\frac{5}{12}\)
TH2:
\(x+\frac{3}{4}=-\frac{1}{3}\)
\(x=-\frac{1}{3}-\frac{3}{4}\)
\(x=\frac{-4-9}{12}\)
\(x=-\frac{13}{12}\)
Vậy x = -5/12 hoặc x = -13/12.
|x + 3/4| - 1/3 = 0
=> |x+3/4| = 1/3
(1) x + 3/4 = 1/3 => x = -5/12
(2) x + 3/4 = -1/3 => x = -13/12
Vậy x =-5/12 hoặc x =-13/12
a) \(\left|x-1,7\right|=2,3\)
\(\Rightarrow\orbr{\begin{cases}x-1,7=2,3\\x-1,7=-2,3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-0,6\end{cases}}\)
b) \(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{3}\\x+\frac{3}{4}=-\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{12}\\x=-\frac{13}{12}\end{cases}}\)
a ) Ta có : \(\left|x\right|=2\frac{1}{3}\)
Đổi : \(2\frac{1}{3}=\frac{7}{3}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{7}{3}\\x=-\frac{7}{3}\end{array}\right.\)
Kết luận : \(x\in\left\{\frac{7}{3};-\frac{7}{3}\right\}\)
b ) \(\left|x\right|=-3\)
Vì : \(x< 0\)
\(\Rightarrow x\) không thõa mãn
Kết luận : \(x\in\left\{\varnothing\right\}\)
c ) \(\left|x\right|=-3,15\)
Vì : \(x< 0\)
\(\Rightarrow x\) không thõa mãn
Kết luận : \(x\in\left\{\varnothing\right\}\)
d ) \(\left|x-1,7\right|=2,3\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1,7=2,3\\x-1,7=-2,3\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-0,6\end{array}\right.\)( thõa mãn )
Kết luận : \(x\in\left\{4;-0,6\right\}\)
e ) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{array}\right.\)
Kết luận \(x\in\left\{-\frac{1}{4};-\frac{5}{4}\right\}\)
\(a,\left|x\right|=2\frac{1}{3}\Rightarrow\left|x\right|=\frac{7}{3}\)
\(\Rightarrow\) \(\begin{cases}x=\frac{7}{3}\\x=\frac{-7}{3}\end{cases}\)
\(b,\left|x\right|=-3\) ( Vì |x| < 0 ) \(\Rightarrow x\in\varnothing\)
\(c,\left|x\right|=-3,15\) (Vì \(\left|x\right|< 0\) ) \(\Rightarrow x\in\varnothing\)
\(d,\left|x-1,7\right|=2,3\)
\(\Rightarrow\) \(\begin{cases}x-1,7=2,3\\x-1,7=-2,3\end{cases}\) \(\Rightarrow\) \(\begin{cases}x=2,3+1,7\\x=-2.3+1,7\end{cases}\) \(\Rightarrow\) \(\begin{cases}x=4\\x=-0,6\end{cases}\)
\(e,\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\) \(\Rightarrow\) \(\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}\) \(\Rightarrow\) \(\begin{cases}x=\frac{1}{2}-\frac{3}{4}=-\frac{1}{4}\\x=-\frac{1}{2}-\frac{3}{4}=-\frac{5}{4}\end{cases}\)
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 ...........
a)Ta có :
\(\frac{4}{5}=\frac{8}{10}\)
1,1=\(\frac{11}{10}\)
Vì \(\frac{8}{10}< \frac{11}{10}\)
»\(\frac{4}{5}< 1,1\)
vậy \(\frac{4}{5}< 1,1\)
a. |x-1,7|=2,3
=> x-1,7=2,3 hoặc x-1,7=-2,3
=> x=2,3+1,7 hoặc x=-2,3+1,7
=> x=4 hoặc x=-1,4
b. |x+3/4|-1/3=0
=> |x-3/4|=1/3
=> x-3/4=1/3 hoặc x-3/4=-1/3
=> x=1/3+3/4 hoặc x=-1/3+3/4
=> x= 13/12 hoặc x=5/12