a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=4.\frac{32}{99}\)

\(=\frac{128}{99}\)

\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=2.\frac{32}{99}\)

\(=\frac{64}{99}\)

\(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)

\(=6\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{203.206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{206}\right)\)

\(=6.\frac{102}{206}\)

\(=\frac{612}{206}\) ( tự rút gọn

=6.3/2.5 +6.3/5.8+...+6.3/203.206

=6(3/2.5+3/5.8+...+3/203.206)

=6(1/2-1/5+1/5-1/8+...+1/203-1/206)

=6[(1/2-1/206)+(1/5-1/5)+(1/8-1/8)+...+(1/203-1/203)]

=6(1/2-1/206)=6(103/206-1/206)=6. 102/206=6. 51/103=306/103

A=6.( 3/2.5+3/5.8+...+3/203.206)

  =6.(1/2-1/5+1/3-1/8+...+1/202-1/206)

  =6.(1/2-1/206)=306/103

A = \(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)

   = \(18.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{203.206}\right)\)

   = \(\frac{18}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{203.206}\right)\)

   = \(6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206}\right)\) 

  = \(6.\left(\frac{1}{2}-\frac{1}{206}\right)\)

  = \(6.\frac{51}{103}\)

  = \(2\frac{100}{103}\)

\(A=\dfrac{18}{2.5}+\dfrac{18}{5.8}+...+\dfrac{18}{203.206}\)

\(A=\dfrac{6.3}{2.5}+\dfrac{6.3}{5.8}+...+\dfrac{6.3}{203.206}\)

\(A=6\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{203.206}\right)\)

\(A=6\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{203}-\dfrac{1}{206}\right)\)

\(A=6\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)

\(A=6.\dfrac{51}{103}\)

\(A=\dfrac{306}{103}\)

18/(2.5) + 18/(5.8) + .... + 18/(203.206)

= 18.[1/(2.5) + 1/(5.8) + .... + 1/(203.206)]

= 18.(1/2 - 1/5 + 1/5 - 1/8 + .... + 1/203 - 1/206)

=18.(1/2 - 1/206)

=18.(51/103)

=918/103

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)

\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)

\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)

\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)

Câu A bạn quên 1/4.5 kìa , với câu D đâu >>>
 

Bài 1:

\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=2.\frac{32}{99}=\frac{64}{99}\)

Bài 2:

a) \(2.4^x-18=110\)

\(\Leftrightarrow2.4^x=128\)

\(\Leftrightarrow4^x=64\)

\(\Leftrightarrow4^x=4^3\Leftrightarrow x=3\)

Vậy x = 3

b) \(\left(\frac{3}{2}x-1\right)^5=1\)

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

\(\Leftrightarrow\frac{3}{2}x=2\)

\(\Leftrightarrow x=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

a) 4/3.5 + 3/5.7 + .... + 4/97.99

= 4( 1/3.5 +1/5.7 + ... + 1/97.99 )

= 4 . 1/2 . 2 ( 1/3.5 +1/5.7 + ... + 1/97.99 )

= 4/2 ( 2/3.5 + 2/5.7 + .... + 2/97.99 )

= 2 ( 5-3/3.5 + 7-5/5.7 + ..... + 99-97/97.99 )

= 2 (5/3.5 - 3/3.5 + 7/5.7 - 5/5.7 + .... + 99/97.99 - 97/97.99 )

= 2 ( 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/97 - 1/99 )

= 2 ( 1/3 -1/99 )

= 2 (33/99 - 1/99 )

= 2 . 32/99

= 32.2/99

=64/99

b) 2. 4^x  - 18 = 110

    2. 4^x = 110 + 18

    2. 4^x = 128 

    4^x = 128 : 2

    4^x = 64

    4^x = 4³

=> x = 3

\(\frac{18}{2.5}+\frac{18}{5.8}+....+\frac{18}{103.106}\)

=\(6\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{103.106}\right)\)

=\(6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{103}-\frac{1}{106}\right)\)

=\(6\left(\frac{1}{2}-\frac{1}{106}\right)\)

=\(6.\frac{26}{53}\)

=\(\frac{156}{53}\)

khó thế ko giải được

609/205