Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
A=\(\dfrac{3\left|x\right|+2}{\left|x\right|-5}=\dfrac{3\left|x\right|-15+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)}{\left|x\right|-5}+\dfrac{17}{\left|x-5\right|}=3+\dfrac{17}{\left|x\right|-5}\)
Để A \(\in\)Z thì \(\left|x\right|-5\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\)
Ta có :
\(\left|x\right|-5=-17\Rightarrow\left|x\right|=-12\left(KTM\right)\)
\(\left|x\right|-5=-1\Rightarrow\left|x\right|=4\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\left|x\right|-5=1\Rightarrow\left|x\right|=6\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
\(\left|x\right|-5=17\Rightarrow\left|x\right|=32\Rightarrow\left[{}\begin{matrix}x=32\\x=-32\end{matrix}\right.\)
Vậy để A \(\in\)Z thì x \(\in\) {-32;-6;-4;4;6;32}
a) \(4,5:\left[\left(\dfrac{9-10}{6}\right)-\dfrac{9}{5}+\dfrac{12}{5}\right]-\dfrac{1}{7}\)
\(=4,5:\left(\dfrac{-1}{6}-\dfrac{-3}{5}\right)-\dfrac{1}{7}\)
=\(4,5:\left(\dfrac{-5+18}{30}\right)-\dfrac{1}{7}\)
=\(4,5:\dfrac{13}{30}-\dfrac{1}{7}\)=\(\dfrac{135}{13}-\dfrac{1}{7}=\dfrac{932}{91}\)
b) \(\dfrac{13}{3}:\left(\dfrac{1}{4}+\dfrac{5}{4}\right)-\dfrac{20}{3}\)
=\(\dfrac{13}{3}.\dfrac{2}{3}-\dfrac{20}{3}\)=\(\dfrac{26}{9}-\dfrac{20}{3}=\dfrac{26}{9}-\dfrac{60}{9}=\dfrac{-34}{9}\)
c) \(5.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+.....+\dfrac{1}{91.94}\right)\)
\(=5.\left[\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\right]\)
\(=5.\left[\dfrac{1}{3}.\left(1-\dfrac{1}{94}\right)\right]\)
=\(5.\left(\dfrac{1}{3}.\dfrac{93}{94}\right)\)
\(=5.\dfrac{31}{94}=\dfrac{155}{94}\)
Chúc bạn học tốt 
\(S=\frac{101}{102}+\frac{1}{1.2.2.3}+\frac{1}{2.3.2.3}+\frac{1}{3.4.2.3}+...+\frac{1}{17.18.2.3}=\frac{101}{102}+\frac{1}{6}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{17.18}\right)\)
Đặt BT trong ngoặc đơn là A
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{18-17}{17.18}=1-\frac{1}{18}=\frac{17}{18}\)
\(S=\frac{101}{120}+\frac{1}{6}.\frac{17}{18}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{1999}{2000}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1999}{2000}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{1999}{2000}\)
\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{1999}{2000}:2\)
\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{1999}{4000}\)
\(\dfrac{1}{n+1}=\dfrac{1}{2}-\dfrac{1999}{4000}\)
\(\dfrac{1}{n+1}=\dfrac{1}{4000}\)
\(\Rightarrow n+1=4000\)
\(n=4000-1\)
\(n=3999\)
Vậy n=3999
a) \(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{5}{3}\)
\(\Rightarrow1-\dfrac{1}{100}+x=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1+\dfrac{1}{100}\)
\(\Rightarrow x=\dfrac{500}{300}-\dfrac{300}{300}+\dfrac{3}{300}\)
\(\Rightarrow x=\dfrac{203}{300}\)
b) \(\left(\dfrac{5}{11.16}+\dfrac{5}{16.21}+...+\dfrac{5}{19.24}\right)-x+\dfrac{1}{3}=\dfrac{7}{3}\)
=>\(\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{19}-\dfrac{1}{24}\right)-x=\dfrac{7}{3}-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{11}-\dfrac{1}{24}-x=2\)
\(\Rightarrow-x=2-\dfrac{1}{11}+\dfrac{1}{24}\)
\(\Rightarrow-x=\dfrac{528}{264}-\dfrac{24}{264}+\dfrac{11}{264}\)
\(\Rightarrow x=\dfrac{515}{264}\)
c) Câu hỏi của Đàm Chu Hữu An - Toán lớp 6 - Học toán với OnlineMath
\(A=2003-\dfrac{1}{2.3}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)
Đặt
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}=\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{20-19}{19.20}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\)
\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)
\(\Rightarrow A=2023-\dfrac{1}{1.2}.B=2023-\dfrac{1}{6}.\dfrac{19}{20}=\)