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21 tháng 8 2017

a)  \(x+\frac{3}{2}=\frac{9}{2}\)

\(\Rightarrow x=\frac{9}{2}-\frac{3}{2}=\frac{6}{2}=3\)

b) \(\frac{1}{3}x+\frac{3}{4}x-75\%=-5\frac{1}{4}\)

\(\Leftrightarrow\frac{4x}{12}+\frac{9x}{12}-\frac{9}{12}=-\frac{63}{12}\)

\(\Leftrightarrow\frac{13x-9}{12}=\frac{-63}{12}\)

\(\Rightarrow13x-9=-63\)

\(13x=-54\Rightarrow x=\frac{-54}{13}\)

21 tháng 8 2017

a) \(x+\frac{3}{2}=\frac{9}{2}\)

\(\Rightarrow x=\frac{9}{2}-\frac{3}{2}\)

\(\Rightarrow x=\frac{6}{2}=3\)

29 tháng 5 2017

a) \(x+\dfrac{3}{2}=\dfrac{9}{2}\)

\(x=\dfrac{9}{2}-\dfrac{3}{2}\)

\(x=3\)

Vậy x = 3 là giá trị cần tìm

b) \(\dfrac{1}{3}x+\dfrac{3}{4}x-75\%=-5\dfrac{1}{4}\)

\(x\left(\dfrac{1}{3}+\dfrac{3}{4}\right)-\dfrac{75}{100}=\dfrac{-19}{4}\)

\(x\left(\dfrac{4}{12}+\dfrac{9}{12}\right)-\dfrac{3}{4}=\dfrac{-19}{4}\)

\(x.\dfrac{13}{12}-\dfrac{3}{4}=\dfrac{-19}{4}\)

\(x.\dfrac{13}{12}=\dfrac{-19}{4}+\dfrac{3}{4}\)

\(x.\dfrac{13}{12}=-4\)

\(x=-4:\dfrac{13}{12}\)

\(x=-\dfrac{48}{13}\)

Vậy..................................

29 tháng 5 2017

a) x + \(\dfrac{3}{2}\)= \(\dfrac{9}{2}\)

=> x= \(\dfrac{9}{2}\)- \(\dfrac{3}{2}\)

=> x=3

Vậy x=3

b) \(\dfrac{1}{3}\)x + \(\dfrac{3}{4}\)x - 75%= \(-5\dfrac{1}{4}\)

=> x.(\(\dfrac{1}{3}\)+ \(\dfrac{3}{4}\)) - \(\dfrac{3}{4}\)= \(\dfrac{-21}{4}\)

=> x. \(\dfrac{13}{12}\)- \(\dfrac{3}{4}\)=\(\dfrac{-21}{4}\)

=> x. \(\dfrac{13}{12}\)= \(\dfrac{-21}{4}\)+ \(\dfrac{3}{4}\)

=> x. \(\dfrac{13}{12}\)= \(\dfrac{-9}{2}\)

=> x= \(\dfrac{-9}{2}\): \(\dfrac{13}{12}\)= \(\dfrac{-9}{2}\). \(\dfrac{12}{13}\)

=> x= \(\dfrac{-54}{13}\)

Vậy x= \(\dfrac{-54}{13}\)

29 tháng 5 2017

Ta có :

\(A=\dfrac{10^{50}+2}{10^{50}-1}=\dfrac{10^{50}-1+3}{10^{50}-1}=\dfrac{10^{50}-1}{10^{50}-1}+\dfrac{3}{10^{50}-1}=1+\dfrac{3}{10^{50}-1}\)

\(B=\dfrac{10^{50}}{10^{50}-3}=\dfrac{10^{50}-3+3}{10^{50}-3}=\dfrac{10^{50}-3}{10^{50}-3}+\dfrac{3}{10^{50}-3}=1+\dfrac{3}{10^{50}-3}\)

\(1+\dfrac{3}{10^{50}-1}< 1+\dfrac{3}{10^{50}-3}\Rightarrow A< B\)

21 tháng 6 2016

\(a.x+9=2\frac{1}{3}\)

    \(x=2\frac{1}{3}-9\)

    \(x=-\frac{20}{3}\)

\(b.\frac{-5}{2}.x=\frac{1}{10}\)

     \(x=\frac{1}{10}:\frac{-5}{2}\)

     \(x=-\frac{1}{25}\)

21 tháng 6 2016

a)

<=> x=\(\frac{7}{3}-9=-\frac{20}{3}\)

b)

<=> x=\(\frac{1}{10}:\frac{-5}{2}=-\frac{2}{25}\)

8 tháng 11 2017

3S= 3/2.5+3/5.8+....+3/17.20

    = (5-2)/2.5+(8-5)/5.8+....+(20-17)/17.20

    = 1/2 - 1/5 + 1/5 - 1/8 + ...... + 1/17 - 1/20 = 1/2 - 1/20 = 9/20

=> S = 3/20

8 tháng 11 2017

\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)

\(S=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(S=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(S=\frac{1}{3}.\frac{9}{20}\)

\(S=\frac{3}{20}\)

21 tháng 6 2016

\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)

\(S=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(S=3.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(S=3.\frac{9}{20}=\frac{27}{20}\)

21 tháng 6 2016

2016-05-06_213811

21 tháng 6 2016

\(a.\frac{2}{3}+\frac{5}{7}=\frac{14}{21}+\frac{15}{21}=\frac{19}{21}\)

\(b.\frac{3}{2}.\frac{4}{5}-\frac{1}{35}=\frac{6}{5}-\frac{1}{35}=\frac{42}{35}-\frac{1}{35}=\frac{41}{35}\)

\(c.\left(\frac{20}{3}-\frac{22}{5}\right):\frac{1}{15}=\left(\frac{100}{15}-\frac{66}{15}\right):\frac{1}{15}=\frac{34}{15}:\frac{1}{15}=\frac{34}{15}x15=34\)

21 tháng 6 2016

a) \(\frac{2}{3}+\frac{5}{7}=\frac{14}{21}+\frac{15}{21}=\frac{14+15}{21}=\frac{29}{21}\)

b)\(\frac{3}{2}.\frac{4}{5}-\frac{13}{5}=\frac{3.4}{2.5}-\frac{13}{5}=\frac{12}{10}-\frac{13}{5}=\frac{6}{5}-\frac{13}{5}=-\frac{7}{5}\)

c)\(\left(\frac{20}{3}-\frac{22}{5}\right):\frac{1}{15}=\left(\frac{100}{15}-\frac{66}{15}\right):\frac{1}{15}=\frac{44}{15}.15=44\)

13 tháng 5 2016

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{50^2}\)

\(A=\frac{1}{1^2}+\frac{1}{2\times2}+\frac{1}{3\times3}+\frac{1}{4\times4}+.....+\frac{1}{50\times50}\)

\(A< \frac{1}{1}+\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+.....+\frac{1}{49\times50}\)

\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)

\(A< 2-\frac{1}{50}\)

\(2-\frac{1}{50}< 2\)

\(\Rightarrow A< 2\)

Chúc bạn học tốtok

 

13 tháng 5 2016

ta có: \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4};...;\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)

=> 

26 tháng 4 2017

Dễ thấy:

\(\dfrac{1}{1^2}=\dfrac{1}{1.1}=\dfrac{1}{1}=1\)

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(.....................\)

\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

Cộng các vế trên với nhau ta được:

\(A< 1+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)

\(\Rightarrow A< 1+\) \(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\Rightarrow A< 1+\left(1-\dfrac{1}{50}\right)\)

\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}\)

\(2-\dfrac{1}{50}< 2\Leftrightarrow A< 2\)

Vậy \(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 2\) (Đpcm)

2 tháng 5 2017

Ta có :

\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+.........+\dfrac{1}{50^2}\)

Ta thấy :

\(\dfrac{1}{1^2}=1\)

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(..............\)

\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..........+\dfrac{1}{49.50}\)

\(A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A< 1+1-\dfrac{1}{50}\)

\(A< 2-\dfrac{1}{50}\)

\(\Rightarrow A< 2\rightarrowđpcm\)

2 tháng 5 2017

đpcm là gì vậy bạn