Với x > 0  

ta có 

x + 1/101 + x  + 2/101 + ... + x + 100/ 101  = 101x 

=> 100x  + ( 1 + 2 + 3 + ... + 100)/101  = 101x 

=>  5050/101 = 101 x - 100x 

=> x = 50 

x < 0 ta có :

   -x - 1/101 - x - 2/101 - ... - x - 100/101 = 101x 

=> - 100x - ( 1 + 2 + .. + 100)/101  = 101x 

=> 5050/101  = -100x - 101x

=> 50          = -201x 

=> x = 

thang Tran trả lời sai, x chỉ có thể lớn hơn 0 thôi, ta có : VT= |x+1/101|+|x+2/101|+|x+3/101|+...+|x+100/101| >= 0

Mà VT=VP =)) VP= 101x >= (lớn hơn hoặc bằng) 0 mà 101 >= 0 =)) x >= 0

<sau đó mới làm giống TH x>0 của bn í>

 SAi vậy mà bn vẫn ak???

Do |x + 1/101| + |x + 2/101| + |x + 3/101| + ... + |x + 100/101| > 0 với mọi x 
mà |x + 1/101| + |x + 2/101| + |x + 3/101| + ... + |x + 100/101| = 101x 
=> x > 0 
Với x > 0 
=> x + 1/101 + x + 2/101 +....+ x + 100/101 = 101x 
<=> x = (1 + 2 + 3 + ... + 100)/101 = 50

Câu 2: 

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)

Có \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge0\)

do đó phương trình ban đầu tương đương với: 

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(\Leftrightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)

\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)

\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+.....+\left|x+\dfrac{100}{101}\right|=101x\left(1\right)\)

VT(1) \(\ge0\) \(\Rightarrow VP\left(1\right)\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=100x+\dfrac{5050}{101}=101x\\ \Rightarrow x=50\)

tim gtln

Ta có: \(\left|x+\frac{1}{101}\right|\ge0\) \(\forall x\)

\(\left|x+\frac{2}{101}\right|\ge0\) \(\forall x\)

\(\left|x+\frac{3}{101}\right|\ge0\) \(\forall x\)

\(............\)

\(\left|x+\frac{100}{101}\right|\ge0\) ∀\(x\)

\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\) \(\forall x\)

\(\Leftrightarrow101x\ge0\)

\(\Leftrightarrow x\ge0\)

\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)

100 hạng tử x 100 số hạng

\(\Leftrightarrow100x+\left(\frac{\left(100+1\right)\cdot100:2}{101}\right)=101x\)

\(\Leftrightarrow100x+\frac{101\cdot50}{101}=101x\)

\(\Leftrightarrow50=101x-100x\)

\(\Rightarrow x=50\)

Vì \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\forall x\)

\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\forall x\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

Từ điều kiện trên ta có :

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(100x+\frac{1+2+...+100}{101}=101x\)

\(101x-100x=\frac{5050}{101}\)

\(x=50\)

Vậy x = 50

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+....+\left|x+\frac{100}{101}\right|=101x\)

\(KĐ:101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)

\(x+\frac{1}{101}+x+\frac{2}{101}+....+x+\frac{100}{101}=101x\)

\(100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)

\(\Rightarrow101-100x=\frac{1+2+....+100}{101}\)

\(x=\frac{\left(1+100\right)\left(100-1+1\right):2}{101}\)

\(x=\frac{101.100:2}{101}\)

\(x=50\)

Vì \(\left|x+\frac{1}{101}\right|+\left|x+\frac{1}{102}\right|+....+\left|x+\frac{100}{101}\right|>0\)

\(\Rightarrow101x>0\)

\(\Rightarrow x>0\)

\(\Rightarrow\left(x+\frac{1}{101}\right)+.....+\left(x+\frac{100}{101}\right)=101x\)

\(\Rightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+....+\frac{100}{101}\right)=101x\)

\(\Rightarrow x=\frac{\left(100+1\right)100:2}{101}\)

\(\Rightarrow x=\frac{50.101}{101}\)

\(\Rightarrow x=50\)

Vậy x = 50

Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)

=> \(101x\ge0\)

=> \(x\ge0\)

=> \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)

=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)

            100 số x                          100 phân số

=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)

=> \(\frac{101.50}{101}=101x-100x\)

=> \(x=50\)

Ta có: \(\left|x+\frac{1}{101}\right|\ge0\); \(\left|x+\frac{2}{101}\right|\) \(\ge0\); ...; \(\left|x+\frac{100}{101}\right|\ge0\)

\(\Rightarrow101x\ge0\)

và \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\); \(\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\); ...; \(\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)

Thay vào đề bài ta đc:

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)

\(\Rightarrow\) \(100x\) + \(\left(\frac{1+2+...+101}{101}\right)=101x\)

\(\Rightarrow100x+101=101x\)

\(\Rightarrow x=101\)

Vậy \(x=101.\)

\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+....+\left|x+\frac{100}{101}\right|\)=101x (1)

điều kiện:101x\(\ge\) 0 \(\Rightarrow\) x\(\ge\) 0

từ (1) \(\Rightarrow\) \(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}\)=101x

\(\Rightarrow\) 100x+(\(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\))=101x

\(\Rightarrow\) 100x+\(\frac{5050}{101}\)=101x

\(\Rightarrow\) \(\frac{5050}{101}\)=101x-100x

\(\Rightarrow\) x=50

k bt mk lm sai hay lm đúng nữa

nếu mk lm sai thì thôi nha!

\(!x+\frac{1}{101}!+!x+\frac{2}{101}!+...+!x+\frac{100}{101}!=101x\) (1)

VT tổng các số không âm => VT>=0 vậy \(VP\ge0\Rightarrow x\ge0\)

với x>=0 biểu thức trong GT tuyệt đối >0 => bỏ dấu trị tuyệt đối biểu thức không đối

do vậy ta có (1) \(\Leftrightarrow\left(x+\frac{1}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)

\(\Leftrightarrow100.x+\left(\frac{1}{101}+...+\frac{100}{101}\right)=101x\)

\(\Leftrightarrow x=\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}=\frac{1}{101}\left(1+2+...+100\right)=\frac{1}{101}\left(\frac{100.101}{2}\right)=50\)

đáp số: x=50

Vì \(VT\ge0\Leftrightarrow VP\ge0\Leftrightarrow101x\ge0\Leftrightarrow x\ge0\)

=>\(\left|x+\frac{1}{101}\right|=x+\frac{1}{101};\left|x+\frac{2}{101}\right|=x+\frac{2}{101};...;\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)

=>\(x+\frac{1}{101}+x+\frac{2}{101}+x+\frac{3}{101}+...+x+\frac{100}{101}=101x\)

=>\(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=100x\)

=>\(100x+50=101x\)

=> x = 50

bài này khó quá ta

mk chịu mất rùi

chúc bn học gioi!

và các bn khác giúp bn này nha'

hihi@@@