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!

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

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

=> \(\frac{101.50}{101}=101x-100x\)

=> \(x=50\)

a ) \(3-4.\left|5-6x\right|=7\)

\(\Leftrightarrow4.\left|5-6x\right|=-4\)

\(\Leftrightarrow\left|5-6x\right|=-1\)

\(\Leftrightarrow\) Không thõa mãn ( vì \(x\ge0\) )

b) Do \(\left|x+2\right|\ge0;\left|x+\frac{3}{5}\right|\ge0;\left|x+\frac{1}{2}\right|\ge0\)

=> \(4x\ge0\)

=> \(x\ge0\)

Lúc này ta có: \(\left(x+2\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{1}{2}\right)=4x\)

=> \(\left(x+x+x\right)+\left(2+\frac{3}{5}+\frac{1}{2}\right)=4x\)

=> \(3x+\frac{31}{10}=4x\)

=> \(4x-3x=\frac{31}{10}\)

=> \(x=\frac{31}{10}\)

Vậy \(x=\frac{31}{10}\)

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

=> \(101x\ge0\)

=> \(x\ge0\)

Lúc này ta có: \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)

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

               100 số x

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

=> \(\frac{101.50}{101}=101x-100x\)

=> \(x=50\)

Vậy x = 50

Bài 1:

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)

Ta thấy:

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\frac{10}{11}=0\)

\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)

Bài 2:

Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)

Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)

a, Vào câu hỏi tương tự nhé

b, Vì \(\hept{\begin{cases}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{cases}\Rightarrow\left|x+3\right|+\left|x+1\right|\ge0\Rightarrow3x\ge0\Rightarrow x\ge0}\)

=> x+3+x+1=3x

=> 2x+4=3x

=>x=4

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

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

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

=> \(2005\ge4-x+10-x+x+990+x+1000+\left|x+101\right|\)

=> \(2005\ge\left|x+101\right|+2004\)

=> \(\left|x+101\right|\le1\)

=> \(x+101\in\left\{-1;0;1\right\}\Rightarrow x\in\left\{-102;-101;-100\right\}\)

d, tương tự b

Câu 1:

Với \(x=11\Rightarrow12=x+1\) ta có: \(x^{17}-12x^{16}+12x^{15}-....+12x-1\)

\(=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-\left(x+1\right)x^{14}+...+\left(x+1\right)x-1\)

\(=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-x^{15}-x^{14}+...-x^3-x^2+x^2+x+1\)

\(=x+1\)

\(=12\)

Câu 2:

Do \(VT>0\Rightarrow VP>0\Rightarrow x>0\Rightarrow\) tất cả các biểu thức dưới dấu trị tuyệt đối đều dương, phương trình trở thành:

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

\(\Leftrightarrow100x+\frac{1+2+3+...+100}{101}=101x\)

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

Câu 3:

\(A=n^3-n+3\left(n^2-1\right)=n\left(n^2-1\right)+3\left(n^2-1\right)\)

\(A=\left(n+3\right)\left(n-1\right)\left(n+1\right)\)

Do n lẻ \(\Rightarrow n=2k+1\)

\(\Rightarrow A=\left(2k+4\right).2k.\left(2k+2\right)=8k.\left(k+1\right)\left(k+2\right)\)

Do \(k\left(k+1\right)\left(k+2\right)\) là tích 3 số nguyên liên tiếp nên chia hết cho 6

\(\Rightarrow A⋮\left(8.6\right)\Rightarrow A⋮48\)