\(a-b=1\Rightarrow b=a-1\)

\(\Rightarrow B=\frac{a-10}{a-1-9}-\frac{2a-a+1}{a+1}=\frac{a-10}{a-10}-\frac{a+1}{a+1}=1-1=0\)

Vậy B = 0

\(a-b=1\Rightarrow b=a-1\)

\(\Rightarrow B=\frac{a-10}{a-1-9}-\frac{2a-a+1}{a+1}=\frac{a-10}{a-10}-\frac{a+1}{a+1}=1-1=0\)

Vậy \(B=0\)

\(-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-....-\frac{1}{1225}\)

\(=-2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{2450}\right)\)

\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\right)\)

\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)

Đặt \(\frac{y+z+1}{2x}=\frac{x+z+2}{2y}=\frac{x+y-3}{2z}=\frac{1}{x+y+z}=k\)

Áp dụng TS DTSBN ta có :

\(k=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{2x+2y+2z}=\frac{2\left(x+y+z\right)}{2\left(x+y+z\right)}=1\)

\(\Rightarrow\hept{\begin{cases}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\end{cases}}\) và \(x+y+z=1\)

\(\Leftrightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\end{cases}}\) và \(x+y+z=1\)

\(\Leftrightarrow\hept{\begin{cases}1+1=3x\\1+2=3y\\1-3=3z\end{cases}\Leftrightarrow\hept{\begin{cases}2=3x\\3=3y\\-2=3z\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{2}{3}\\y=1\\z=-\frac{2}{3}\end{cases}}}\)

Vậy \(x=\frac{2}{3};y=1;z=-\frac{2}{3}\)

ko biết

ai trả lời giùm mình,mình k cho

bn rảnh quá

Ta có : A = |2x+2|+|2x-2013|

           A = |2x+2|+|2013-2x| \(\ge\)2x+2+2013-2x=2015

    Dấu ''='' xảy ra khi \(\hept{\begin{cases}2x+2\ge0\\2013-2x\ge0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}2x\ge2\\2x\le2013\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le1006\end{cases}}\)\(\left(x\in Z\right)\)\(\Leftrightarrow1\le x\le1006\)

Vậy để A = |2x+2|+|2x-2013| đạt GTNN là 2015 thì \(1\le x\le1006\)

Hok tốt

ta có

A = |2x + 2| + |2x - 2013|

 |2x + 2| \(\ge\) \(2x+2\)\(\forall\)  \(x\in Z\)

  |2x - 2013|  \(\ge\) \(2013-2x\)   \(\forall\) \(x\in Z\)

\(\Rightarrow\text{​​}\) A = |2x + 2| + |2x - 2013|  \(\ge\)\(2x+2\)  +   \(2013-2x\)  \(=\)       \(2015\)         \(\forall\)\(x\in Z\)

dấu bằng xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}2x+2\ge0\\2013-2x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}2x\ge-2\\x\le1006\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\le1006\end{cases}}}\)

vậy min A=2015  \(\Leftrightarrow\)  \(-1\le x\le1006\)