x=2009 dễ mà

mk làm câu c cho nó dễ

c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010

=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010

=1-1/x+1=2009/2010

=1/x+1=1-2009/2010

=1/x+1=1/2010

=) x+1=2010

x         =2010-1

x         =2009

Bạn gõ lại đề đi :v

Đọc chả hiểu đề gì cả ... đề k có x

Mà phía dưới có cái đáp số x= ... là sao ??

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

x=\(\frac{4}{13}\)

\(1+\frac{1}{3}+\frac{1}{6}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)\right]=4\)

\(\Leftrightarrow1+2\left(\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=4\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{4-1}{2}=\frac{3}{2}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2}-\frac{3}{2}=-1\)

\(\Leftrightarrow x=-1+1=-2\)

Vậy x = -2 

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}=\frac{1}{1993}\)

\(\Leftrightarrow x+1=1993\)

\(\Leftrightarrow x=1992\)

a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

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

=> \(A=\frac{9}{10}\)

b/ \(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)

=> \(A=1+\frac{7}{n-5}\)

Để A nguyên => 7 chia hết cho n-5 => n-5=(-7; -1; 1; 7)

=> n=(-2; 4, 6, 8)

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{6}{7}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{6}{7}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{6}{7}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{6}{7}=\frac{1}{7}\)

\(\Rightarrow x+1=7\)

\(\Rightarrow x=7-1=6\)

vậy x = 6

\(\frac{1}{1.2}+\frac{1}{2.3}+............+\frac{1}{x\left(x+1\right)}=\frac{6}{7}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...........-\frac{1}{x}-\frac{1}{x+1}=\frac{6}{7}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{6}{7}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{7}\)

\(\Rightarrow x+1=7\)

\(\Rightarrow x=6\)

Vậy x = 6

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}\div2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{4032}\)

\(\Rightarrow x+1=4032\Rightarrow x=4031\)

Vậy \(x=4031\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2016}\)

=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2015}{2016}\)

=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.x+1}\right)=\frac{2015}{2016}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}:2\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2032}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{2032}\)

=> \(\frac{1}{x+1}=\frac{1}{2032}\)

Vì 1 = 1

=> x + 1 = 2032

=> x = 2032 - 1

=> x = 2031

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2015}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(1-\frac{1}{x+1}=1-\frac{2015}{2016}\)

\(\frac{1}{x+1}=\frac{1}{2016}\)

\(x=2016-1\)

\(\Rightarrow x=2015\)

a)  \(=\frac{1}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}.\frac{6.6}{5.7}=\frac{6}{2.7}=\frac{3}{7}\)

B) \(=\frac{70}{11}+\frac{1}{9}-\frac{37}{11}-\frac{1}{9}=\left(\frac{70}{11}-\frac{37}{11}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)=\frac{33}{11}+0=3\)

BÀI 2:

A) \(\Leftrightarrow\frac{7}{2}x-\frac{x}{2}+\frac{2x}{2}=\frac{7}{2}.\frac{5}{6}\)

\(\Leftrightarrow\frac{7x-x+2x}{2}=\frac{35}{12}\)

\(\Leftrightarrow\frac{8x}{2}=\frac{35}{12}\)

\(\Leftrightarrow8x.12=35.2\Leftrightarrow96x=70\Leftrightarrow x=\frac{70}{96}=\frac{35}{48}\)

b) \(\left(x-\frac{3}{1.2}\right)+\left(x-\frac{3}{2.3}\right)+...+\left(x-\frac{3}{99.100}\right)=1\)

\(x-\frac{3}{1.2}+x-\frac{3}{2.3}+....x+\frac{3}{99.100}=1\)

\(\Leftrightarrow\left(x+x+x+...+x\right)-3\left(\frac{1}{1.2}+\frac{1}{1.3}+....+\frac{1}{99.100}\right)=1\)

ngoặc 1 có 99 số hạng x

\(\Leftrightarrow99x-3\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=1\)

\(\Leftrightarrow99x-3\left(1-\frac{1}{100}\right)=1\)

\(\Leftrightarrow99x-3.\frac{99}{100}=1\)

\(\Leftrightarrow99x=1+\frac{3.99}{100}\)

\(\Leftrightarrow99x=\frac{397}{100}\)

\(\Leftrightarrow x=\frac{397}{100.99}=\frac{397}{9900}\)

bó tay