Ta chứng minh bài toán \(a_1\le a_2\le...\le a_n\) thỏa mãn \(a_1+a_2+...+a_n=0;\left|a_1\right|+\left|a_2\right|+...+\left|a_n\right|=1\) thì \(a_n-a_1=\frac{2}{n}\) 

Từ điều kiện trên ta có \(k\in N\) sao cho \(a_1\le a_2\le...a_k\le0\le a_{k+1}\le...\le a_n\)

\(\Rightarrow\hept{\begin{cases}\left(a_1+a_2+...+a_k\right)+\left(a_{k+1}+...+a_n\right)=0\\-\left(a_1+a_2+...+a_k\right)+\left(a_{k+1}+...+a_n\right)=\frac{1}{2}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a_1+a_2+...+a_k=-\frac{1}{2}\\a_{k+1}+...+a_n=\frac{1}{2}\end{cases}}\). Mà 

\(a_1\le a_2\le...\le a_k\Rightarrow a_1\le-\frac{1}{2k};a_{k+1}\le...\le a_n\Rightarrow a_n\ge\frac{1}{2k}\)

\(\Rightarrow a_n-a_1\ge\frac{1}{2k}+\frac{1}{2\left(n-k\right)}=\frac{n}{2k\left(n-k\right)}\ge\frac{n}{2\left(\frac{k+n-k}{2}\right)^2}=\frac{2}{n}\)

Áp dụng vào bài chính theo giải thiết ta có: 

\(\hept{\begin{cases}\frac{x_1}{2013}+\frac{x_2}{2013}+...+\frac{x_{192}}{2013}=0\\\left|\frac{x_1}{2013}\right|+\left|\frac{x_2}{2013}\right|+...+\left|\frac{x_{192}}{2013}\right|=0\end{cases}}\)

\(\Rightarrow\frac{x_{192}}{2013}-\frac{x_1}{2013}\ge\frac{2}{192}\Rightarrow x_{192}-x_1\ge\frac{2013}{96}\)

\(\frac{x}{3}=\frac{y}{4}=k\Rightarrow x=3k\) ; \(y=4k\)

Ta có : \(x.y=192\Rightarrow3k.4k=192\)

  \(12k^2=192\Rightarrow k^2=16\Rightarrow\orbr{\begin{cases}k=4\\k=-4\end{cases}}\)

Với \(k=4\Rightarrow x=4.3=12\); \(y=4.4=16\)

Với \(k=-4\Rightarrow x=-4.3=-12\); \(y=-4.4=-16\)

Vậy x = 12 hoặc -12  ; y = 16 hoặc -16 

đề bài là tìm x;y;z

\(\frac{x}{3}=\frac{y}{4}=>\frac{x}{3}.\frac{y}{4}=\frac{y}{4}.\frac{y}{4}=>\frac{xy}{12}=\frac{y^2}{16}=\frac{192}{12}=16\)

\(=>y^2=16.16=256\)

\(=>y=-16;16\)

\(=>\orbr{\begin{cases}y=-16=>x=192:\left(-16\right)=-12\\y=16=>x=192:16=12\end{cases}}\)

Vậy x=12;y=16

x=-12;y=-16

192 x 81 + 192 x 18 + 192

= 192 x ( 81 + 18 + 1 )

= 192 x 100

= 19 200.

192 x 81 + 192 x 18 + 192

= 192 x ( 81 + 18 ) + 192

= 192 x 99 + 192

= 19008 + 192

= 19200

k nhé Nguyễn Việt Thái

Seira Nguyễn tinh như thế vẫn chưa phải là tính thuận tiện đâu nhé

két quả là:125/96

2/3+ 2/6+ 2/12+ 2/24+ 2/48+ 2/192= 2/3+ 1/3+ 1/6+1/12+1/24+1/96

                                                        =64/96+32/96+16/96+8/96+4/96+1/96

                                                        =125/96

\(=\frac{125}{96}\)

   ( 192 - 37 + 85 ) - ( 85 + 192 )

= 192 - 37 + 85 - 85 - 192

= 192 - 192 + 85 - 85 - 37

= ( 192 - 192 ) + ( 85 - 85 ) - 37

= 0 + 0 - 37

= - 37

(192-37+85)-(85+192)

=192-37+85-85-192

=(192-192)+(85-85)-37

=-37

Ta thấy tất cả các phân số đều có mẫu chung là 192

=> \(\frac{128+64+32+16+8+4+2}{192}\)

= \(\frac{254}{192}\)= \(\frac{127}{96}\)

2/3+2/6+2/12+2/24+2/48+2/96+2/192

=2/3+1/3+1/6+1/12+1/24+1/48+1/96

=127/96

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{29}{96}+\frac{2}{192}\)

\(=\frac{2}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

\(=\frac{254}{192}=\frac{127}{96}\)

192* = 1926 vì 1+9+2+6=18 , 18 chia hết cho 9

\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(2C=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}\)

\(2C-C=\frac{4}{3}-\frac{2}{192}\)

\(C=\frac{127}{96}\)

\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(C=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

\(C=\frac{254}{192}=\frac{127}{96}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

=\(\frac{2}{3}+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

    Ta xét phần trong ngoặc :

\(\frac{1}{2}+\frac{1}{4}=1-\frac{1}{4}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=1-\frac{1}{8}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=1-\frac{1}{64}\)

=\(\frac{63}{64}\)

=> \(1+\frac{63}{64}=\frac{127}{96}\)

Vậy tổng trên có kết quả là \(\frac{127}{96}\)

k nha