\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{64}{13}\)

\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{64}{13}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{13}\div5\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{65}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{64}{65}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{65}\)

\(\Rightarrow x+1=65\Rightarrow x=65-1=64\)

\(\text{Vậy }x=64\)

\(2xy+x+2y=13\\ \Rightarrow2xy+x+2y+1-1=13\\ \Rightarrow\left(2xy+2y\right)+\left(x+1\right)=13+1\\ \Rightarrow2y\left(x+1\right)+\left(x+1\right)=14\\ \Rightarrow\left(x+1\right)\left(2y+1\right)=14\\ \Rightarrow\left(x+1\right);\left(2y+1\right)\inƯ\left(14\right)\\ \Rightarrow\left(x+1\right);\left(2y+1\right)\in\left\{-14;-7;-2;-1;1;2;7;14\right\}\)

Vì \(x,y\in N\Rightarrow\left(x;y\right)=\left(0;\dfrac{13}{2}\right),\left(1;3\right),\left(6;\dfrac{1}{2}\right),\left(13;0\right)\)

Vậy \(\left(x;y\right)=\left(0;\dfrac{13}{2}\right),\left(1;3\right),\left(6;\dfrac{1}{2}\right),\left(13;0\right)\)

sxdcc

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)

\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)

\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)

\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)

\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)

\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)

\(\Leftrightarrow x+\frac{103}{50}=5\)

\(\Leftrightarrow x=5-\frac{103}{50}\)

\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)

\(\Leftrightarrow x=\frac{147}{50}\)

Số x cần tìm là : 99

Ta có : \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)

\(\Rightarrow5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{20}.\frac{1}{5}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)

=> x + 1 = 100

=> x = 99

Ta có  \(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)

   =>  \(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)

   =>    \(5\left(1-\frac{1}{x+1}\right)=\frac{99}{20}\)

   =>   \(1-\frac{1}{x+1}=\frac{99}{100}\)

  =>   \(\frac{1}{x+1}=\frac{1}{100}\)

 =>   x+1 = 100

=>  x = 99

Cai phan 1+3+5+...+99 chac em biet lam roi phai ko? Con 3/1.2+3/2.3+3/3.4+...+3/99.100 thi em cu tach lam sao cho tro thanh dang ban dau thi lam . Anh phai nghi roi !~ Neu chieu anh ranh ranh thi len giai tiep . BYE BYE

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}+4x=1+3+5+...+99\)

\(3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+4x=\left(1+99\right)+\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)\)\(3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)+4x=100+100+100+...+100\)\(3\left(1-\frac{1}{100}\right)+4x=100.25\)

\(3.\frac{99}{100}+4x=2500\)

\(\frac{297}{100}+4x=2500\)

\(4x=2500-\frac{297}{100}\)

\(4x=2500-2,97\)

\(4x=2497,03\)

\(x=624,2575\)

\(x=2497,03:4\)

\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{9998}{9999}\)

\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{9998}{9999}\)

\(\Leftrightarrow5\left(1-\frac{1}{x+1}\right)=\frac{9998}{9999}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{9998}{9999}\div5\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{9998}{49995}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{9998}{49995}=\frac{39997}{49995}\)

\(\Leftrightarrow x=\frac{9998}{39997}\)