ai biết ko 1phần3+5phần8-7phần12

Đặt A=\(\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+........+\frac{1}{4000}\)

A=\(1+\left(1+\frac{3999}{2}\right)+\left(1+\frac{3998}{3}\right)+........+\left(1+\frac{1}{4000}\right)\)

A=\(\frac{4001}{4001}+\frac{4001}{2}+\frac{4001}{3}+...........+\frac{4001}{4000}\)

A=\(4001.\left(\frac{1}{2}+\frac{1}{3}+........+\frac{1}{4000}+\frac{1}{4001}\right)\)

=>\(y=\frac{4001.\left(\frac{1}{2}+\frac{1}{3}+........+\frac{1}{4001}\right)}{\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{4001}}\)

=>\(y=4001\)

A=[(3999/2+1)+(3998/3+1)+...+(1/4000+1)+1]/(1/2+1/3+...+1/4001)

A=(4001/2+4001/3+...+4001/4001)/(1/2+1/3+...+1/4001)

A=[4001(1/2+1/3+...+1/4001)]/(1/2+1/3+...+1/4001)

A=4001

Vậy A=4001

! ) A = (3999 /2 +1 ) + ( 3998/ 3 + 1 ) + ( 3997 / 4 + 1 ) +...+ ( 1/ 4000 + 1 ) + 1

 (Ta lấy 4000/1 = 4000  rải đều  1, 1 ,1 cho 3999 phân số  và dư lại 1  = 4001/4001 )

= 4001 /2 + 4001 / 3 + 4001 /4 + ...+ 4001 /4000 + 4001 / 4001

= 4001 ( 1/2 + 1/3 + 1/4 +..+ 1/ 4001 )  vay A: B = 4001

Đặt P = ... ( biểu thức đề bài ) 

Nhận xét: Với \(k\inℕ^∗\) ta có: 

\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)

\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)

bài 1:

\(\frac{6}{11}+\frac{1}{3}+\frac{5}{11}\)

\(=\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{1}{3}\)

\(=\frac{11}{11}+\frac{1}{3}=1+\frac{1}{3}=\frac{3}{3}+\frac{1}{3}=\frac{4}{3}\)

bài 2:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{12}\right)\)

\(=\frac{11}{20}+\frac{1}{4}=\frac{11}{20}+\frac{5}{20}=\frac{15}{20}=\frac{3}{4}\)

bài 3: 

a) \(\frac{3}{2}\cdot\frac{4}{5}\cdot\frac{2}{3}=\left(\frac{3}{2}\cdot\frac{2}{3}\right)\cdot\frac{4}{5}=1\cdot\frac{4}{5}=\frac{4}{5}\)

b) \(\frac{6}{7}\cdot\frac{5}{3}\cdot\frac{7}{6}=\left(\frac{6}{7}\cdot\frac{7}{6}\right)\cdot\frac{5}{3}=1\cdot\frac{5}{3}=\frac{5}{3}\)

bài 4: 

a) \(\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{4}\cdot\frac{2}{5}=\frac{2}{5}\cdot\left(\frac{1}{4}+\frac{3}{4}\right)=\frac{2}{5}\cdot1=\frac{2}{5}\)

b) \(\frac{6}{11}:\frac{2}{3}+\frac{5}{11}:\frac{2}{3}=\left(\frac{6}{11}+\frac{5}{11}\right):\frac{2}{3}=1:\frac{2}{3}=\frac{3}{2}\)

Bài 1: 

  6/11 + 1/3 + 5/11 

= ( 6/11 + 5/11) + 1/3

= 11/11 + 1/3

= 1 + 1/3 

= 3/3 +1/3  

= 4/3 

Bài 2: 

1/2 + 1/6 + 1/12 + 1/20 

= ( 1/2 + 1/6 + 1/12 ) + 1/20 

= ( 6/12 + 2/12 + 1/12 ) + 1/20 

=9/12 + 1/20 

= 3/4 +1/20  

= 15/20 + 1/20 

= 16/20 = 4/5

 Bài 3:

a) \(\frac{3}{2}\times\frac{4}{5}\times\frac{2}{3}\) \(=\left(\frac{3}{2}\times\frac{2}{3}\right)\times\frac{4}{5}\)\(=1\times\frac{4}{5}=\frac{4}{5}\) 

b)  \(\frac{6}{7}\times\left(\frac{5}{3}\times\frac{7}{6}\right)\) \(=\frac{6}{7}\times\frac{35}{18}\)\(=\frac{1\times5}{7\times3}=\frac{5}{21}\)   

Bài 4:

a) 2/5  x 1/4 + 3/4 x 2/5 

= 2/5 x ( 1/4  + 3/4) 

 = 2/5 x  1 

= 2/5

 b) 6/11 : 2/3 +5/11 : 2/3 

=  ( 6/11 + 5/11) x 3/2 

= 11/11 x 3/2  

= 1 x 3/2  

=  3/2  

....  

=5/6

l.ik.e minh nha Đặng Khánh Linh