khó was ak

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Lập bảng xét dấu là cách dài nhất nhưng dễ nhất

 2005 = |x - 4| + |x - 10| + |x + 101| + |x + 990| + |x + 1000|

=> 2005 = x - 4 + x - 10 + x + 101 + x + 990 + x + 1000

=> 2005 = 5x + (-4 - 10 + 101 + 990 + 1000) 

=> 2005 = 5x + 2077

=> 5x = 2005 - 2077 = -72

=> x = -72/5 

 2005 = |x - 4| + |x - 10| + |x + 101| + |x + 990| + |x + 1000|

=> 2005 = x - 4 + x - 10 + x + 101 + x + 990 + x + 1000

=> 2005 = 5x + (-4 - 10 + 101 + 990 + 1000) 

=> 2005 = 5x + 2077

=> 5x = 2005 - 2077 = -72

=> x = -72/5 

@trần như

Mình thử lại thấy kết quả ko thỏa mãn

Mình thấy: (2005= x - 4 + x - 10 + x + 101+ x + 990 + x + 1000) sai

 2005 = |x - 4| + |x - 10| + |x + 101| + |x + 990| + |x + 1000|

=> 2005 = x - 4 + x - 10 + x + 101 + x + 990 + x + 1000

=> 2005 = 5x + (-4 - 10 + 101 + 990 + 1000) 

=> 2005 = 5x + 2077

=> 5x = 2005 - 2077 = -72

=> x = -72/5 

@tranthithao tran

Mình thấy kết quả \(\frac{-72}{5}\)thử lại nó không đúng.

Ta co /x-4/+/x-10/+/x+101/+/x+990/+/x+1000/= /x+101/+/x-4/+/x-10/+/x+990/+/x+1000/= /x+101/+/x-4/+/x-10/+/x+990/+/x+1000/

>= /x+101/+/x-4+x-10+x+990+x+1000/=/x+101/+2013

=> 2013>=/x+101/+2013=.> /x+101/=<0

ma /x+101/>=0nen/x+101/=0=.>x=-101

Sai rồi bạn ơi

a, \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

=>\(2^{x-1}+\frac{5}{2}.2^{x-1}=\frac{7}{32}\)

=>\(2^{x-1}\left(1+\frac{5}{2}\right)=\frac{7}{32}\)

=>\(2^{x-1}\cdot\frac{7}{2}=\frac{7}{32}\)

=>\(2^{x-1}=\frac{1}{16}=\frac{1}{2^4}=2^{-4}\)

=>x-1=-4

=>x=-5

b, |x - 4| + |x - 10| + |x + 101| + |x + 990| + |x + 1000| = |4-x|+|10-x|+|x+101|+|x+990|+|x+1000|

Ta có: \(\left|4-x\right|\ge4-x;\left|10-x\right|\ge10-x;\left|x+990\right|\ge x+990;\left|x+1000\right|\ge x+1000\)

\(\Rightarrow\left|4-x\right|+\left|10-x\right|+\left|x+990\right|+\left|x+1000\right|\ge4-x+10-x+x+990+x+1000\)

\(\Rightarrow\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\ge2004+\left|x+101\right|\)

\(\Rightarrow2005\ge2004+\left|x+101\right|\)

\(\Rightarrow\left|x+1\right|\le1\)

\(\Rightarrow-1\le x+101\le1\)

\(\Rightarrow-102\le x\le-100\)

Vì \(x\in Z\)

\(\Rightarrow x\in\left\{-102;-101;-100\right\}\)

bài a nhầm 2 dòng cuối

=>x-1=-4

=>x=-3

Bài a bạn ST làm sai

2004=(x-x-x+x+x)+(4+10+101+990+1000)

còn tiếp theo thì mk ko biết

Ta có: 

\(\left|x-4\right|+\left|x-10\right|+\left|x+101\right|+\left|990\right|+\left|x+1000\right|\)

\(=\left|x+101\right|+\left|x-4\right|+\left|x-10\right|+\left|x+990\right|+\left|x+1000\right|\)

\(\ge\left|x+101\right|+\left|x-4+x-10+x+990+x+1000\right|x+101\left|+\right|2004\)

\(\Rightarrow2004\ge\left|x+101\right|+2004\Rightarrow\left|x+101\right|\le0\)

Mà: \(\left|x+101\right|\ge0\)nên \(\left|x+101\right|=0\Rightarrow x=-101\)

Vậy: \(x=-101\)