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15 tháng 5 2022

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15 tháng 5 2022

kb với mik để mik giải tiếp cho nha

b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)

\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)

\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)

=>x=3/11+20/55=3/11+4/11=7/11

c: \(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-2}{98}-1\right)+\left(\dfrac{x-5}{95}-1\right)=\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)

\(\Leftrightarrow x-100=1\)

hay x=101

5 tháng 4 2018

A+B = (1.99+2.98+3.97+...+99.1)+(1.101+2.102+3.103+...+99.199)

A+B = (1.99+1.101)+(2.98+2.102)+(3.97+3.103)+...+(99.1+99.199)

A+B = 1(99+101) + 2(98+102) + 3(97.103)+...+99(1+199)

A+B = 1.200 + 2.200 + 3.200 +...+ 99.200

A+B = 200.(1+2+3+...+200)

A+B = 200.4950

A+B = 990000

20 tháng 4 2017

A + B = ( 1 . 99 + 2 . 98 + 3 . 97 + ... + 99 . 1 ) + ( 1 . 101 + 2 . 102 + 3 . 103 + ... + 99 . 199 )

A + B = 99 . ( 1 + 199 ) + 98 . ( 2 + 198 ) + 97 . ( 3 + 197 ) + ... + 2 . ( 102 + 98 ) + 1 . ( 99 + 101 ) 

A + B = 99 . 200 + 98 . 200 + 97 . 200 + ... + 2 . 200 + 1 . 200

A + B = ( 99 + 98 + 97 + ... + 2 + 1 ) . 200

A + B = 4950 . 200

A + B = 990000

5 tháng 3 2019

Ta có:
$(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$

$\Leftrightarrow \frac{1}{100}\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=\frac{1}{10}\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$

$\Leftrightarrow \left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$

Đặt $A=\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}$

$\Rightarrow A=\left ( 1+\frac{1}{2}+...+\frac{1}{10} \right )+\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{100} \right )-\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{100} \right )-\left (\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right )$

$\Rightarrow A=\left ( 1+\frac{1}{2}+...+\frac{1}{10} \right )-\left (\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right )$

$\Rightarrow A=\frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}$

Thay vào phương trình, ta có:

$\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )$

$\Leftrightarrow x=10$

20 tháng 2 2018

(100/1.101 + 100/2.102 + 100/3.103 +....+100/10.110) . x

= (10/1.11 + 10/2.12 + 10/100.110 )10

=>(1+1/2+1/3+...+1/10-1/101-...-1/110)x

=(1+1/2+1/3+...+1/10+1/11+...+1/100-1/11-...-1/100-1/101-...-1/110)10 =>(1+1/2+1/3+...+1/10-1/101-...-1/110)x

=(1+1/2+1/3+...+1/10-1/101-...-1/110)10 =>x=10