\(1-\frac{303030}{313131}+\frac{616161}{626262}-1+\frac{929292}{939393}-1=\left(1-\frac{30}{31}\right)+\left(\frac{61}{62}-1\right)+\left(\frac{92}{93}-1\right)\)

\(=\frac{1}{31}-\frac{1}{62}-\frac{1}{93}=\frac{1}{186}\)

=> \(\frac{\left|x\right|}{186}=\frac{1}{186}\)=> |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

phải làm mới đc **** chứ !

x = 184

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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

\(\frac{\left|x\right|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)+\left(\frac{929291}{939393}-1\right)\)

<=> \(\frac{\left|x\right|}{186}=-\frac{303030}{313131}+\frac{616161}{626262}+\frac{929292}{939393}+1-1-1\)

<=> \(\frac{\left|x\right|}{186}=-\frac{30}{31}+\frac{61}{62}+\frac{92}{93}+1-1-1\)

<=> \(\frac{\left|x\right|}{186}=\frac{61}{62}+\frac{92}{93}-1-\frac{30}{31}\)

<=> \(\frac{\left|x\right|}{186}=-\frac{1.31}{31}+\frac{61}{62}+\frac{92}{91}-\frac{30}{31}\)

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

<=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

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

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.