giá trị x thỏa mãn
\(\frac{|x|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)\)là x=
\(\frac{\left|x\right|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)+\left(\frac{929292}{939393}-1\right)\)
tìm x
\(1-\frac{303030}{313131}+\frac{616161}{626262}-1+\frac{929292}{939393}-1=\left(1-\frac{30}{31}\right)+\left(\frac{61}{62}-1\right)+\left(\frac{92}{93}-1\right)\)
\(=\frac{1}{31}-\frac{1}{62}-\frac{1}{93}=\frac{1}{186}\)
=> \(\frac{\left|x\right|}{186}=\frac{1}{186}\)=> |x| = 1 => x = 1 hoặc x = - 1
Vậy...............
303030/313131=30/31(1)
616161/626262=61/62(2)
929292/939393=92/92(3)
từ 1,2,3=>tự làm
\(\frac{\text{\x\}}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)+\left(\frac{929292}{939393}-1\right)\)
tim x nguyen
ma cai dau \x\ la gia tri tuyet doi
giai ra giup minh voi
xim may ban trnh bay chi tiet duoc khong
cam on rat nhieu
\(\dfrac{x}{186}=\left(1-\dfrac{303030}{313131}\right)+\left(\dfrac{616161}{626262}-1\right)+\left(\dfrac{929292}{939393}-1\right)\)
x/186 = (1 - 303030/313131) + (616161/626262 - 1) + (929292/939393 - 1)
x/186 = 1 - 30/31 + 61/62 - 1 + 92/93 - 1
x/186 = 61/62 - 30/31 + 92/93 - 1
x/186 = 1/62 - 1/93
x/186 = 1/186
x = 1
|x|/186=(1-303030/313131)+(616161/626262-1)+(929292/939393-1)
\(\frac{\left|x\right|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)+\left(\frac{929291}{939393}-1\right)\)
<=> \(\frac{\left|x\right|}{186}=-\frac{303030}{313131}+\frac{616161}{626262}+\frac{929292}{939393}+1-1-1\)
<=> \(\frac{\left|x\right|}{186}=-\frac{30}{31}+\frac{61}{62}+\frac{92}{93}+1-1-1\)
<=> \(\frac{\left|x\right|}{186}=\frac{61}{62}+\frac{92}{93}-1-\frac{30}{31}\)
<=> \(\frac{\left|x\right|}{186}=-\frac{1.31}{31}+\frac{61}{62}+\frac{92}{91}-\frac{30}{31}\)
Lấy MSC là 168, ta có:
<=> \(\frac{\left|x\right|}{186}=\frac{-186}{186}+\frac{183}{186}+\frac{184}{186}-\frac{180}{186}\)
<=> \(\frac{\left|x\right|}{186}=\frac{-186+183+184-180}{186}\)
<=> \(\frac{\left|x\right|}{186}=\frac{1}{186}\)
<=> |x| = 1
<=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
12) X/186=(1-303030/313131)+(616161/626262-1)+(929292/939393-1)
Tìm x biết:
a) | x+ 1/3| = 5/ 6
b) 11/ 13- ( 5/ 42- x) = -( 15/ 28 - 11/ 13)
c) | x| / 186= ( 1- 303030/ 313131) + ( 616161/ 626262- 1)
a ) | x + 1/3 | = 5/6
+) x + 1/3 = 5/6 hay x + 1/3 = -5/6
x = 5/6 - 1/3 x = -5/6 - 1/3
x = 1/2 x =-7/6
b ) 11/13 - ( 5/42 - x ) = - ( 15/28-11/13 )
11/13 - ( 5/42 - x ) = 113/364
( 5/42 - x ) = 11/13 - 113/364
( 5/42 - x ) = 15/28
x = 5/42 - 15/28
x = -5/12
c ) | x |/186 = ( 1 - 303030 / 313131 ) + ( 616161 / 626262 - 1 )
| x | / 186 = 1/62
| x | = 1/62 . 186
| x | = 3
x = 3 hoặc -3
số các giá trị của x thỏa mãn :
\(\frac{\left|x-7\right|}{\left|x-4\right|}=\frac{\left|x-1\right|}{\left|x-4\right|}\)là bao nhiêu?
số giá trị của x thỏa mãn \(\frac{\left|4-x\right|+\left|x+2\right|}{\left|x+5\right|+\left|x-3\right|}=-\frac{1}{2}\) là
Xét tử \(\left|4-x\right|+\left|x+2\right|\ge0\)
Xét mẫu \(\left|x+5\right|+\left|x-3\right|\ge0\)
Do đó \(\frac{\left|4-x\right|+\left|x+2\right|}{\left|x+5\right|+\left|x-3\right|}\ge0\)
Nhưng đề bài cho \(\frac{\left|4-x\right|+\left|x+2\right|}{\left|x+5\right|+\left|x-3\right|}=-\frac{1}{2}<0\) nên không có giá trị nào của x thỏa mãn.
Giá trị của x thỏa mãn đẳng thức
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|=100x\)