Sửa đề: 3 bài 5 điểm

a: 

b: Trung bình cộng là:

\(\dfrac{3\cdot3+4\cdot3+5\cdot3+6\cdot4+7\cdot10+8\cdot7}{30}=6,2\)

3/2 . x + ( 5/3 - 3/2) : 2/3 = 5/3

3/2.x + 1/6 : 2/3 = 5/3

3/2.x + 1/4 = 5/3

3/2.x = 5/3 - 1/4

3/2.x=17/12

x= 17/12 : 3/2

x= 17/18

Vậy...

Bài 2:

4/5x7 + 4/7x9 + 4/9x11 +...+4/17x19

= 2(2/5.7 + 2/7.9 + 2/9.11+...+ 2/17/19)

= 2( 1/5 - 1/7 + 1/7 -1/9 + 1/9 -1/11 +...+ 1/17 - 1/19)

= 2( 1/5- 1/19)

= 2 . 14/95

= 28/95

Trả lời:

Bài 1 

\(\frac{3}{2}\times x+\left(\frac{5}{3}-\frac{3}{2}\right)\div\frac{2}{3}=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{6}\div\frac{2}{3}=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{6}\times\frac{3}{2}=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{4}=\frac{5}{3}\)

\(\Leftrightarrow\frac{1}{6}\times x=\frac{17}{12}\)

\(\Leftrightarrow x=\frac{17}{2}\)

Vậy \(x=\frac{17}{2}\)

Trả lời

Bài 1

\(\frac{3}{2}\times x+\left(\frac{5}{3}-\frac{3}{2}\right)\div\frac{2}{3}=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{6}\div\frac{2}{3}=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{6}\times\frac{3}{2}=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}\times x+\frac{1}{4}=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}\times x=\frac{17}{12}\)

\(\Leftrightarrow x=\frac{17}{18}\)

Vậy \(x=\frac{17}{18}\)

Bài 2 

\(\frac{4}{5\times7}+\frac{4}{7\times9}+\frac{4}{9\times11}+...+\frac{4}{17\times19}\)

\(=2\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{17\times19}\right)\)

\(=2\times\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{19}\right)\)

\(=2\times\left(\frac{1}{5}-\frac{1}{19}\right)\)

\(=2\times\frac{14}{95}\)

\(=\frac{28}{95}\)

mik gửi lầm link đây nhé https://olm.vn/hoi-dap/detail/1459935110952.html?auto=1

Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{203}{3^{100}}< 3\)

\(A< \frac{3}{4}\left(đpcm\right)\)

CMR: \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{100}{4^{100}}< \frac{4}{9}\)

Dạng tổng quát: CMR: \(\frac{1}{k}+\frac{2}{k^2}+\frac{3}{k^3}+\frac{4}{k^4}+...+\frac{n}{k^n}< \frac{k}{\left(k-1\right)^2}\)(k;n \(\in\) N*; k > 1)

Cách 1:

7/5 + 4/3 + 7/5 + 4/3 = (7/5 + 7/5) + (4/3+4/3) = 14/5 + 8/3 = 42/15 + 40/15 = 82/15

Cách 2:

7/5 + 4/3 + 7/5 + 4/3 :

7/5 và 4/3 đều có mẫu chung = 15, suy ra:

7/5=21/15

4/3=20/15, suy ra:

21/15 + 20/15 + 21/15 + 20/15 = 82/15

lm đầu tiên nha. nhớ đấy

C1

7/5+4/3+7/5+4/3

=41/15+41/15

=82/15

C2

7/5+4/3+7/5+4/3

=(7/5+7/5)+(4/3+4/3)

=14/5+8/3

=82/15

fkldmdkgmjfk,d/ckjxffm

Bài 1:

a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)

= \(\dfrac{29}{10}\)

b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)

= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)

= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)

= \(\dfrac{19}{20}\)

c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

= \(\dfrac{29}{6}\)

Bài 2:

   3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\) 

= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)

= \(\dfrac{28}{5}\)

b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)

= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)

= \(\dfrac{43}{34}\)

Bài 3:

a, 800 kg 5g > 8,005 kg

b, 9 ha 4 dam² < 9\(\dfrac{5}{100}\) ha

c, 5m 5mm > 5,0005 m

d, 250 kg > \(\dfrac{1}{5}\) tấn

Bài 1 :

chứng minh A = 2 + 2^2 + 2^3 + ........... + 2^2009 + 2^2010 chia hết 42

ta thấy 42 = 2 x 3 x  7

A chia hết 42 suy ra A phải chia hết cho 2;3;7

mà ta thấy tổng trên chia hết cho 2 suy ra A chia hết cho 2  (1)

số số hạng ở tổng A là : ( 2010 - 1 ) : 1 + 1 = 2010 ( số )

ta chia tổng trên thành các nhóm mỗi nhóm 2 số ta được số nhóm là : 2010 : 2 = 1005 ( nhóm )

suy ra A = ( 2 + 2^2 ) + ( 2^3 + 2^4 ) + ...............+ ( 2^2009 + 2^2010 )

A = 2 x ( 1 + 2 ) + 2^3 x ( 1 + 2 ) + ................. + 2^2009 x ( 1 + 2 )

A = 2 x 3 + 2^3 x 3 + ............. + 2^2009 x 3 

A = 3 x ( 2 + 2^3 + ........... + 2^2009 ) chia hết cho 3 

suy ra A chia hết cho 3 ( 2 )

ta chia nhóm trên thành các nhóm mỗi nhóm 3 số ta có số nhóm là : 2010 : 3 = 670 ( nhóm )

suy ra A = ( 2 + 2^2 + 2^3 ) + ( 2^4 + 2^5 + 2^6 ) + ................. + ( 2^2008 + 2^2009 + 2^2010 )

A = 2 x ( 1 + 2 + 2^2 ) + 2^4 x ( 1 + 2 + 2^2 ) + .................. + 2^2008 x ( 1 + 2 + 2^2 )

A = 2 x ( 1 + 2 + 4 ) + 2^4 x ( 1 + 2 + 4 ) + ................ + 2^2008 x ( 1 + 2 + 4 )

A = 2 x 7 + 2^4 x 7 + ............. + 2^2008 x 7

A = 7 x ( 1 + 2^4 + ........ + 2^2008 ) chia hết cho 7 

suy ra A chia hết cho 7 (3)

từ (1) ; (2) và (3) suy ra A chia hết cho 2;3;7 

suy ra A chia hết cho 42 ( điều phải chứng minh )