x + y = 2

=>(x - 1) + (y - 1) = 0

=> x - 1 đối y - 1

=> (x - 1)(y - 1) 

=> (x - 1)(y - 1) ≤ 0

=> xy - x - y + 1 ≤ 0

=> xy - (x + y) + 1 ≤ 0

=> xy - 2 + 1 ≤ 0

=> xy - 1 ≤ 0

=> xy < 1 (đpcm)

= \(\frac{183}{100}\). \(\frac{11}{5}\)+ \(\frac{-168}{25}\). \(\frac{11}{5}\)- \(\frac{11}{100}\). \(\frac{11}{5}\)- ( - 1 )

=\(\frac{11}{5}\). ( \(\frac{183}{100}\)+ \(\frac{-168}{25}\)-\(\frac{11}{100}\)) + 1

=\(\frac{11}{5}\). ( - 5 ) +1

= - 11 + 1

= - 10

Ax1000=1830.11/5-672.22-110.11/5+1000

=(1830.11-672.22.5-110.11+5000)/5

=(1830.11-6720.11-110.11+5000)/5

=(1830.11-6830.11+5000)/5

=(-5000.11+5000)/5

=-5000.10/5

=1000.10

Có ^xOy + ^xOz = ^yOz

thì tia Ox nằm giữa hai tia Oy và Oz. 

Xin lỗi nhưng theo mình nhớ thì có 5 cách thì phải

Đặt C =\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

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

              \(=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow C=\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\div2\)

a) \(1,25:0,8=\frac{3}{8}:0,2x\)

\(\Rightarrow\frac{3}{8}:0,2x=\frac{25}{16}\)

\(\Rightarrow0,2x=\frac{3}{8}:\frac{25}{16}=\frac{6}{25}\)

\(\Rightarrow x=\frac{6}{25}:\frac{1}{5}=\frac{6}{5}\)

b) \(1,8:1,3=\left(-2,7\right):5x\)

\(\Rightarrow\frac{-27}{10}:5x=\frac{18}{13}\)

\(\Rightarrow5x=\frac{-27}{10}:\frac{18}{13}=\frac{-39}{20}\)

\(\Rightarrow x=\frac{-39}{100}\)

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2017}\)

\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{x}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2017}\)

\(\Rightarrow-\frac{1}{x+3}=\frac{1}{2017}\)

\(\Rightarrow x+3=-2017\)

\(\Rightarrow x=-2020\)

Để A < 0

mà x + 7 > x - 3 

=> \(\hept{\begin{cases}x+7>0\\x-3< 0\end{cases}}\)

=> \(\hept{\begin{cases}x>-7\\x< 3\end{cases}}\)

=> - 7 < x < 3

Vậy với mọi x thỏa mãn - 7 < x < 3 thì A luôn âm

Cảm ơn nhé Nguyễn Chí Thành