Ta có : \(\dfrac{1}{1794}\)>\(\dfrac{1}{1795^2}\)

\(\dfrac{1}{1794}\)>\(\dfrac{1}{1796^2}\)

\(\dfrac{1}{1794}\)>\(\dfrac{1}{1797^2}\)

.....................

\(\dfrac{1}{1794}\)>\(\dfrac{1}{2016^2}\)

\(\dfrac{1}{1794}\)>\(\dfrac{1}{2017^2}\)

\(\Leftrightarrow\)\(\dfrac{1}{1794}\)>\(\dfrac{1}{1795^2}\)+\(\dfrac{1}{1796^2}\)+\(\dfrac{1}{1797^2}\)+. . .+\(\dfrac{1}{2016^2}\)+\(\dfrac{1}{2017^2}\)

Chúc bạn học tốt

a) Vì \(\dfrac{x+5}{3}\)= \(\dfrac{x-6}{7}\) nên 7(x+5) = 3(x-6)

=> 7x+ 35 = 3x - 18

7x - 3x = -18 -35

4x = -53

x = -53:4

x = \(\dfrac{-53}{4}\)

\(\frac{1}{1975^2}+\frac{1}{1976^2}+...+\frac{1}{2017^2}< \frac{1}{1974.1975}+\frac{1}{1975.1976}+...+\frac{1}{2016.2017}\)

\(=\frac{1}{1974}-\frac{1}{1975}+\frac{1}{1975}-\frac{1}{1976}+...+\frac{1}{2016}-\frac{1}{2017}=\frac{1}{1974}-\frac{1}{2017}< \frac{1}{1974}\)

Lời giải:

$A=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}$

$B=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}$

Vì $20^{10}-1> 20^{10}-3$

$\Rightarrow \frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}$

$\Rightarrow 1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}$

$\Rightarrow A< B$

Giải:

Ta có:

A=20¹⁰+1/20¹⁰-1

A=20¹⁰-1+2/20¹⁰-1

A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-1<2/20¹⁰-3 nên A<B

Chúc bạn học tốt!

a, sai đề

b, \(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Rightarrow\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\) ( nhân cả 2 vế với \(\dfrac{1}{2}\) )

\(\Rightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\Rightarrow x+1=18\Rightarrow x=17\)

Vậy x = 17

Câu a thiếu đề rồi bạn ơi mik giải câu b đây:

\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)

\(2\left(\dfrac{1}{6}-\dfrac{1}{x+2}\right)=\dfrac{2}{9}\)

\(\dfrac{1}{6}-\dfrac{1}{x+2}=\dfrac{2}{9}:2\)

\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)

\(\dfrac{1}{x+1}=\dfrac{1}{18}\)

\(\Rightarrow x+1=18\Rightarrow x=17\)

Vậy x = 17

\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

\(\dfrac{2}{20^{10}-1}>\dfrac{2}{20^{10}-3}\Leftrightarrow A>B\)

Bài 1: Ta có:

\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)

\(=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)

\(=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)

\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+...+\dfrac{1}{26}-\dfrac{1}{32}\right)\)

\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)

Vậy \(M=\dfrac{5}{16}\)

Bài 2: Ta có:

\(A=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+...+\dfrac{1}{210}\)

\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+...+\dfrac{1}{14.15}\)

\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

\(=\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{1}{10}\)

Vậy \(A=\dfrac{1}{10}\)

Giải:

\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}.\)

\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+\dfrac{5}{208}.\)

\(M=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+\dfrac{5}{13.16}.\)

\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}\left[\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(\dfrac{1}{10}-\dfrac{1}{10}\right)+\left(\dfrac{1}{13}-\dfrac{1}{13}\right)+\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\right].\)

\(M=\dfrac{5}{3}\left[0+0+0+\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\right]\)

\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}\left(\dfrac{4}{16}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}.\dfrac{3}{16}.\)

\(M=\dfrac{15}{48}=\dfrac{5}{16}.\)

M=\(\dfrac{5}{28}+\dfrac{1}{14}+\dfrac{1}{26}+\dfrac{5}{208}\)

M=\((\dfrac{5}{28}+\dfrac{1}{14})+\left(\dfrac{1}{26}+\dfrac{5}{208}\right)\)

M=\(\dfrac{1}{4}+\dfrac{1}{16}\)

M=\(\dfrac{5}{16}\)

A=\(\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+...+\dfrac{1}{13\times14}+\dfrac{1}{14\times15}\)

A=\(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

Sau khi giản ước các phân số cho nhau Ta có:

A=\(\dfrac{1}{6}-\dfrac{1}{15}\)

A=\(\dfrac{1}{10}\)