a,Ta có \(\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{1-\frac{2}{3}-\frac{1}{2}}-\frac{\frac{3}{5}-\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}-\frac{6}{7}-\frac{6}{11}}\)

\(=\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{2.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)}-\frac{3.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}{6.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}\)

=\(\frac{1}{2}-\frac{3}{6}=\frac{1}{2}-\frac{1}{2}=0\)

Vậy giá trị biểu thức bằng 0

b, Mình không hiểu cho lắm ạ , nếu ko phiền xin xem lại đầu bài ạ

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  

....  

C = 1/3 + -3/4 + 3/5 + 1/57 + -1/36 + 1/15 + -2/9

C = ( 1/3 + 1/57 ) + ( -3/4 + -1/36 ) + ( 3/5 + 1/15 ) + -2/9 

C = ( 19/57 + 1/57 ) + ( -27/36 + -1/36 ) + ( 9/15 + 1/15 ) + -2/9 

C = 20/57 + -28/36 + 10/15 + -2/9 

C = 20/57 + -7/9 + 2/3 + -2/9

C = ( 20/57 + 2/3 ) + ( -7/9 + -2/9 )

C = 58/57 + -1 

C = 1/57

D = 1/2 + -1/5 + -5/7 + 1/6 + -3/35 + 1/3 + 1/41

D = ( 1/2 + 1/3 + 1/6 ) + ( -1/5 + -5/7 +-3/35 ) + 1/41

D = ( 3/6 + 2/6 + 1/6 ) + ( -7/35 + -25/35 + -3/35 ) + 1/41

D = 1 + -1 + 1/41

D = 1/41

E = -1/2 + 3/5 + -1/9 + 1/127 + -7/18 + 4/35 + 2/7 

E = ( -1/2 + -1/9 + -7/18 ) + ( 3/5 + 4/35 ) + 1/127 + 2/7

E = ( -9/18 + -2/18 + -7/18 ) + ( 21/35 + 4/35 ) + 1/127 + 2/7

E = -1 + 5/7 + 1/257 + 2/7 

E = -1 + ( 5/7 + 2/7 ) + 1/127

E = -1 + 1 + 1/127

E = 1/127

\(C=\frac{1}{3}+\frac{-3}{4}+\frac{3}{5}+\frac{1}{57}+\frac{-1}{36}+\frac{1}{15}+\frac{-2}{9}.\)

\(C=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{1}{36}+\frac{2}{9}\right)+\frac{1}{57}\)

\(C=1-1+\frac{1}{57}\)

\(C=\frac{1}{57}\)

D=1/2+-1/5+-5/7+1/6+-3/35+1/3+1/41

D=(1/2+1/6+1/3)-(1/5+3/35+5/7)+1/41

D=1-1+1/41

D=1/41

E=-1/2+3/5+-1/9+1/127+-7/18+4/35+2/7

E=(3/5+4/35+2/7)-(1/2+1/9+7/18)+1/127

E=1-1+1/127

E=1/127

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{481}{280}\)

Bạn phải giải ra 

Đặt P = ... ( biểu thức đề bài ) 

Nhận xét: Với \(k\inℕ^∗\) ta có: 

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

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

\(\frac{1}{5}\cdot\frac{1}{7}+\frac{4}{5}+\frac{1}{5}\cdot\frac{12}{7}-1-\frac{1}{5}\cdot\frac{6}{7}\)

\(=\frac{1}{5}\cdot\left(\frac{1}{7}+\frac{12}{7}-\frac{6}{7}\right)-\left(1-\frac{4}{5}\right)\)

\(=\frac{1}{5}.1-\frac{1}{5}\)

\(=\frac{1}{5}-\frac{1}{5}=0\)