\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}.\)

\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)

\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)

\(\Leftrightarrow x\cdot\frac{5}{3}=15\)

\(\Leftrightarrow x=15:\frac{5}{3}\)

\(\Leftrightarrow x=15\cdot\frac{3}{5}\)

\(\Leftrightarrow x=9.\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{1}-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)

\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)

\(\Rightarrow x.\frac{5}{3}=14+1=15\)

\(\Rightarrow x=15:\frac{5}{3}=9\)

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{1}-\frac{1}{2018}\)

\(=\frac{2018}{2018}-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b) \(\left(x\cdot\frac{5}{3}-1\right):9=3\frac{1}{2}:2,25\)

\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{7}{2}:\frac{9}{4}\)

\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{14}{9}\)

\(x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)

\(x\cdot\frac{5}{3}-1=14\)

\(x\cdot\frac{5}{3}=14+1\)

\(x\cdot\frac{5}{3}=15\)

\(x=15:\frac{5}{3}\)

\(x=9\)

Vậy \(x=9\)

5.(1/5+1/17)-(2/5+2/17+9/15+12/68)

=5.22/85-22/17

=22/17-22/17

=0

Ta có : \(5\cdot\left(\frac{1}{5}+\frac{1}{17}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{9}{15}+\frac{12}{68}\right)\)

\(=\) \(5\cdot\frac{1}{5}+5\cdot\frac{1}{17}-\left(\frac{2}{5}+\frac{2}{17}+\frac{3}{5}+\frac{3}{17}\right)\)

\(=\) \(1+\frac{5}{17}-\left[\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{2}{17}+\frac{3}{17}\right)\right]\)

\(=\) \(1+\frac{5}{17}-\left(1+\frac{5}{17}\right)\)

\(=\) \(1+\frac{5}{17}-1-\frac{5}{17}\)

\(=\)\(0\)

     Vậy ... 

                  Tk ủng hộ mk nha các bn ❣❣ C.ơn nhiều ^^

1 1/9 x 1 1/10 x 1 1/11 x ... x 1 1/2011

=10/9 x 11/10 x 12/11 x ... x 2012/2011

khử

còn 2012/9

=\(\frac{10}{9}\)x\(\frac{11}{10}\)x\(\frac{12}{11}\)x.........x\(\frac{2012}{2011}\)

=\(\frac{2012}{9}\)

\(1\frac{1}{9}.1\frac{1}{10}.1\frac{1}{11}...1\frac{1}{2011}\)

= \(\frac{10}{9}.\frac{11}{10}.\frac{12}{11}...\frac{2012}{2011}\)

= \(\frac{2012}{9}\)

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1-\dfrac{1}{2017}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}=\dfrac{1}{2017}\)

Giải:

\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right).\left(1-\dfrac{1}{2017}\right)\)

\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}\)

\(\Leftrightarrow A=\dfrac{1.2...201.2016}{2.3...2016.2017}\)

\(\Leftrightarrow A=\dfrac{1.2.3...2015.2016}{2017.2.3...2015.2016.}\)

Rút gọ cả tử và mẫu với 2.3...2015.2016, ta được:

\(A=\dfrac{1}{2017}\)

Vậy \(A=\dfrac{1}{2017}\).

Chúc bạn học tốt!

Giải:

\(A=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2015}{2016}\cdot\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2017}\)

Vậy, A= \(\dfrac{1}{2017}\)

tôi đã thử lòng các bạn nhưng ko có ai trả lời thì tớ giải cho nhé.

bài làm:  Đặt \(\frac{x}{1998}=\frac{y}{1999}=\frac{z}{2000}=k\Rightarrow\)x =1998k   ; y =1999k   ; z =2000k

ta có : \(\left(x-z\right)^3=\left(1999k-2000k\right)^3\)  = \(\left[k\cdot\left(1999-2000\right)\right]^3\)= \(k^3\cdot\left(-8\right)\)                                         (1)

\(8\cdot\left(x-y\right)^2\cdot\left(y-z\right)\) = \(8\cdot\left(1998k-1999k\right)^2\cdot\left(1999k-2000k\right)\) 

                                                  = \(8\cdot\left[k\cdot\left(1999-2000\right)\right]^2\cdot\left[k\cdot\left(1999-2000\right)\right]\)

                                                 = \(8\cdot k^2\cdot1\cdot k\cdot\left(-1\right)=k^3\cdot\left(-8\right)\)                                                                        (2)

từ (1)và (2) \(\Rightarrow\left(x-z\right)^3=8\cdot\left(x-y\right)^2\cdot\left(y-z\right)\)  

Ta có \(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5.\)

\(\Rightarrow\frac{2}{3}-\frac{1}{3}.x+\frac{1}{3}.\frac{3}{2}-\frac{1}{2}.2x-\frac{1}{2}=5\)

\(\Rightarrow\frac{2}{3}-\frac{x}{3}+\frac{1}{2}-x-\frac{1}{2}=5\)

\(\Rightarrow\frac{4}{6}-\frac{2x}{6}+\frac{3}{6}-\frac{6x}{6}-\frac{3}{6}=\frac{30}{6}\)

\(\Rightarrow4-2x+3-6x-3=30\)

\(\Rightarrow4-8x=30\)

\(\Rightarrow-8x=26\)

\(\Rightarrow x=\frac{26}{-8}=-\frac{13}{4}\)

Vậy \(x=-\frac{13}{4}\)

x bằng 80 nhé

Ta có \(\left(1-\frac{1}{97}\right)\times\left(1-\frac{1}{98}\right)\times.....\times\left(1-\frac{1}{1000}\right).\)

\(=\frac{97-1}{97}\times\frac{98-1}{98}\times.....\times\frac{1000-1}{1000}\)

\(=\frac{96}{97}\times\frac{97}{98}\times....\times\frac{999}{1000}\) (rút gọn hết )

\(=\frac{96}{1000}\)

\(=\frac{12}{125}\)