\(\frac{121212}{161616}-\left(\frac{151515}{323232}-x\right)=2\)

=> \(\frac{3}{4}-\left(\frac{15}{32}-x\right)=2\)

=> \(\frac{15}{32}-x=\frac{3}{4}-2\)

=> \(\frac{15}{32}-x=-\frac{5}{4}\)

=> \(x=\frac{15}{32}-\frac{-5}{4}=\frac{15}{32}+\frac{5}{4}=\frac{55}{32}\)

b) \(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}+\frac{x}{90}=\frac{9}{5}\)

=> \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}+\frac{x}{6\cdot7}+\frac{x}{7\cdot8}+\frac{x}{8\cdot9}+\frac{x}{9\cdot10}=\frac{9}{5}\)

=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{9}-\frac{x}{10}=\frac{9}{5}\)

=> \(\frac{x}{1}-\frac{x}{10}=\frac{9}{5}\)

=> \(\frac{10x-x}{10}=\frac{9}{5}\)

=> \(\frac{9x}{10}=\frac{9}{5}\)

=> \(\frac{9x}{10}=\frac{9\cdot2}{5\cdot2}=\frac{18}{10}\)

=> x = 2

1  cặp có giá trị là:

\(\frac{1}{11}\)+\(\frac{1}{25}\)=\(\frac{36}{275}\)

Có các phân số là;

(25-11):1+1=15(phân số)

Có các cặp là :

15 :2=7(CẶP ,DƯ 1 CẶP)

1 CẶP DƯ ĐÓ LÀ:

\(\frac{36}{275}\):2=\(\frac{36}{550}\)=\(\frac{18}{275}\)

Các cặp có tổng là:

\(\frac{36}{275}\).7=\(\frac{252}{275}\)

Tổng số đó là:

\(\frac{252}{275}\)+\(\frac{18}{275}\)=\(\frac{270}{275}\)=\(\frac{54}{55}\)

Phân số \(\frac{54}{55}\)lớn   hơn phân số \(\frac{47}{60}\)vì

\(\frac{54}{55}\)và \(\frac{47}{60}\)=\(\frac{3240}{3300}\)và \(\frac{2585}{3300}\)

\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{25}\)

\(=\left(\frac{1}{11}+\frac{1}{12}\right)+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+\frac{1}{24}+\frac{1}{25}\right)\)

\(\frac{1}{11}+\frac{1}{12}>\frac{1}{12}+\frac{1}{12}=\frac{2}{12}=\frac{10}{60}\)

\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}>\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{3}{15}=\frac{12}{60}\)

\(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}=\frac{5}{20}=\frac{15}{60}\)

\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+\frac{1}{24}+\frac{1}{25}>\frac{1}{25}+\frac{1}{25}+\frac{1}{25}+\frac{1}{25}+\frac{1}{25}=\frac{5}{25}=\frac{1}{5}=\frac{12}{60}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{25}>\frac{10}{60}+\frac{12}{60}+\frac{15}{60}+\frac{12}{60}=\frac{49}{60}\)

Mà \(\frac{49}{60}>\frac{47}{60}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{25}>\frac{47}{60}\left(đpcm\right)\)

12 000 - ( 1 500 . 2 + 1 800 . 3 + 1 800 . 2 . 3)

= 12 000 - ( 8400 + 108000

= 12 000 - 19200

= -7200

=12000- (3000+  (1800.3+2)

=  12000-   3000+ 9000

=   12000-  12000

=0

a) Ta có : 3 > 2 và 300 > 200

\(\Rightarrow3^{300}>2^{200}\)

b) Ta có : 1000 > 999

\(\Rightarrow5^{1000}>5^{999}\)

c) Ta có : \(243^5=\left(3^5\right)^5=3^{25}\)

                 \(3.243^5=3.\left(3^5\right)^5=3.3^{25}=3^{26}\)

                 \(3.27^8=3.\left(3^3\right)^8=3.3^{24}=3^{25}\)

mà 25 = 25 < 26

\(\Rightarrow3^{25}=3^{25}< 3^{26}\)

\(\Rightarrow243^5=3.27^8< 3.243^5\)

d) Ta có : \(125^5=\left(5^3\right)^5=5^{15}\)

                 \(25^7=\left(5^2\right)^7=5^{14}\)

mà 15 > 14

\(\Rightarrow5^{15}>5^{14}\)

\(\Rightarrow125^5>25^7\)

a, vì 199<200;201>200 nên hai kq liền nhau

Mà 199x201 tận cùng là 9 200.200 tận cùng là 0 => A<B

tương tự b

vô lí! đoạn cuối phải là 2/196.199 chứ

??????????

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{99}}\)

\(3A-A=1-\frac{1}{3^{100}}\)

\(2A=1-\frac{1}{3^{100}}\)

\(A=\frac{1-\frac{1}{3^{100}}}{2}\)

Ta có: 4²⁰⁰ = (4²)¹⁰⁰ = 16¹⁰⁰

2⁴⁰⁰ = (2⁴)¹⁰⁰ = 16¹⁰⁰

Vì 16¹⁰⁰ = 16¹⁰⁰ nên 4²⁰⁰ = 2⁴⁰⁰

4^200=2^400