\(\text{Ta có: }\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{x.\left(x+1\right)}=\frac{13}{90}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{13}{90}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

=> x = 17

a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)

⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)

⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)

⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)

⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)

⇒ \(\frac{1}{x+1}=\frac{1}{18}\)

⇒ x + 1 = 18

⇒ x = 17

Vậy x = 17

b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)

⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)

⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)

⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)

⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)

⇒ \(\frac{1}{x+3}=\frac{1}{148}\)

⇒ x + 3 = 148

⇒ x = 145

Vậy x = 145

Theo đề suy ra

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)

=> \(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)

\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)

=>x+1=30

=>x=29

\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)x<\frac{13}{7}\)

\(\left(1-\frac{1}{7}\right).x<\frac{13}{7}\)

\(\frac{6}{7}.x<\frac{13}{7}\Leftrightarrow6x<13\Leftrightarrow x<2,1\left(6\right)\)

x nguyên dương => x thuộc {1;2}

Vậy tập hợp có 2 phần tử

vay tap hop co 2 phan tu

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2016}\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2016}\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{1}{2016}\Leftrightarrow\frac{x+1-5}{5\left(x+1\right)}=\frac{1}{2016}\Leftrightarrow\frac{x-4}{5x+5}=\frac{1}{2016}\Rightarrow2016\left(x-4\right)=5x+5\Leftrightarrow2016x-8064=5x+5\)còn lại tự giải

Bài dễ mà !

1/3.4+1/4.5+1/5.6+.....+1/x(x+1)=3/10

1/3-1/4+1/4-1/5+1/5-........-1/x+1/x-1/x+1=3/10

=>1/3-1/x+1=3/10

   1/x+1=3/10-1/3=1/30

=>x+1=30

   x=30-1

   x=29

Ta có :

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)

=>\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)

=>\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)

=>\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)

=>\(\frac{1}{x+1}=\frac{1}{30}\)

=>\(x+1=30\)

=>\(x=30-1\)

=>\(x=29\)

Vậy \(x=29\)

b  \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{19}{100}\)

=>\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)

=>\(\frac{1}{5}-\frac{1}{x+1}\)\(=\frac{19}{100}\)

=>\(\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)

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

=> x+1 =100

=>x=99

b) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{19}{100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{19}{100}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)

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

\(\Rightarrow x+1=100\)

\(\Rightarrow x=99\)

c) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{49}{99}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{49}{99}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{49}{99}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{49}{99}\)

\(\Rightarrow\frac{1}{x+2}=\frac{50}{99}\)

\(\Rightarrow50.\left(x+2\right)=99\)

\(\Rightarrow x+2=\frac{99}{50}\)

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

d) Ta có : 6 = 1.6 = 2.3 = (-2) . (-3)

Lâp bảng xét 6 trường hợp: 

Vậy các cặp (x,y) \(\inℤ\)thỏa mãn là : (0;4) ; (1; 4) ; (-2 ; 0)

e) \(x^2-3xy+3y-x=1\)

\(\Rightarrow x\left(x-3y\right)+3y-x=1\)

\(\Rightarrow x\left(x-3y\right)-\left(x-3y\right)=1\)

\(\Rightarrow\left(x-3y\right)\left(x-1\right)=1\)

Lại có : 1 = 1.1 = (-1) . (-1)

Lập bảng xét các trường hợp : 

Vậy các cặp(x,y) thỏa mãn là : \(\left(2;\frac{1}{3}\right);\left(0;\frac{1}{3}\right)\)