Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Theo đề bài cho => \(4-\frac{5}{6-\frac{7}{8-\frac{9}{10}}}=4-\frac{5}{x}\)
=> \(x=6-\frac{7}{8-\frac{9}{10}}\)
Vậy x=...
Nếu thế mk đúng thì ủng hộ nha
Ta có:\(2-\frac{3}{4-\frac{5}{6-\frac{7}{8-\frac{9}{10}}}}=2-\frac{3}{4-\frac{5}{6-\frac{7}{\frac{71}{10}}}}=2-\frac{3}{4-\frac{5}{6-\frac{70}{71}}}\)
\(=2-\frac{3}{4-\frac{5}{\frac{356}{71}}}\)
=>x=\(\frac{356}{71}\)
\(\frac{2x+1}{3}=\frac{x-5}{2}\)
\(\Rightarrow2\left(2x+1\right)=3\left(x-5\right)\)
\(\Rightarrow4x+2=3x-15\)
\(\Rightarrow4x-3x=-15-2\)
\(\Rightarrow x=-17\)
Ta có:
\(\frac{2x+1}{3}=\frac{x-5}{2}\)
\(\Rightarrow\left(2x+1\right)2=\left(x-5\right)3\)
\(\Rightarrow4x+2=3x-15\)
\(\Rightarrow4x-3x=-15-2\)
\(\Rightarrow x=-17\)
Vậy x = -17
\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)
=>\(\frac{\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)}{99}=-4\)
=> (x+1)+(x+2)+(x+3)+(x+4)=-4.99=-396
=>4x+10=-396
4x=-406
x=-406:4=-101,5
\(\frac{3}{5}.x-\frac{1}{2}.x=-\frac{8}{20}\)
\(\left(\frac{3}{5}-\frac{1}{2}\right).x=-\frac{2}{5}\)
\(\frac{1}{10}.x=-\frac{2}{5}\)
\(x=\left(-\frac{2}{5}\right):\frac{1}{10}\)
\(x=-4\)
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
\(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
\(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{3}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}\)
\(=\frac{5}{6}\)
2E=1+\(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2003}}\)
2E-E=1-\(\frac{1}{2^{2004}}\)
E=\(\frac{1}{2^{2004}}\)
Ủng hộ mk nha
\(\frac{16}{5}.\frac{5}{11}+\frac{17}{5}.\frac{5}{11}-\frac{1}{2}\)
\(=\frac{5}{11.}\left(\frac{16}{5}+\frac{17}{5}\right)-\frac{1}{2}\)
\(=\frac{5}{11}.\frac{33}{5}-\frac{1}{2}\)
\(=3-\frac{1}{2}\)
\(=\frac{5}{2}\)
~Hok tốt~
\(\frac{x-1}{3}=-\frac{2}{5}\)
\(\Rightarrow\frac{5\left(x-1\right)}{15}=\frac{-6}{15}\)
\(\Rightarrow5\left(x-1\right)=-6\)
\(\Rightarrow x-1=-\frac{6}{5}\)
\(\Rightarrow x=-\frac{6}{5}+1\)
\(\Rightarrow x=-\frac{1}{5}\)
Vậy \(x=-\frac{1}{5}\)
\(\frac{x-1}{3}=-\frac{2}{5}\)
\(\Leftrightarrow\left(x-1\right).5=3.-2\)
\(\Leftrightarrow5x-5=-6\)
\(\Leftrightarrow x=-\frac{1}{5}\)