Ghi đầy đủ nha

bn có thể ghi rõ ràng đc ko?

a)\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{3}{7}\) 

=\(\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\dfrac{5}{13}\) 

=\(0+\dfrac{5}{13}\) 

=\(\dfrac{5}{13}\)

A = 2/1*5 + 2/5*9 + ... + 2/101*105

   = 1/2(4/1*5 + 4/5*9 + ... + 4/101*105)

   = 1/2(1 - 1/5 + 1/5 - 1/9 + ... + 1/101 - 1/105)

   = 1/2(1 - 1/105)

   = 1/2 * 104/105 = 52/105

Sửa câu b. Phân số thứ 2 phải là 4/5*8

B = 4/2*5 + 4/5*8 + ... + 4/47*50

   = 4/3(3/2*5 + 3/5*8 + ... + 3/47*50)

   = 4/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)

   = 4/3(1/2 - 1/50)

   = 4/3 * 24/50 = 16/25

\(P=\frac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)\(=\frac{\left(1+4\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)...\left(20^2+1\right)\left(\cdot22^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)\left(6^2+1\right)\left(8^2+1\right)\left(10^2+1\right)\left(12^2+1\right)...\left(22^2+1\right)\left(24^2+1\right)}\)

\(=\frac{1+4}{\left(2^2+1\right)\left(24^2+1\right)}=\frac{5}{5\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)

cái bước tách ra bn nhân lại là có kết quả y chang, VD:

\(\left(5^4+4\right)=\left(4^2+1\right)\left(6^2+1\right)=629\)

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)

yeu cau cua bai la gi?

rut gon da thuc

có \(a^4+4=a^4+4a^2+4-4a^2\)

                  =\(\left(a^2-2\right)^2-4a^2\)

                    =\(\left(a^2-2-2a\right)\left(a^2-2+2a\right)\)

                    =\(\left[\left(a-1\right)^2-1\right]\left[\left(a+1\right)^2-1\right]\)

thay vào biểu thức A ta có;

A=\(\frac{\left[\left(1-1\right)^2-1\right]\left[\left(1+1\right)^2-1\right]...\left[\left(21-1\right)^2-1\right]\left[\left(21+1\right)^2-1\right]}{\left[\left(3-1\right)^2-1\right]\left[\left(3+1\right)^2-1\right]...\left[\left(23-1\right)^2-1\right]\left[\left(23+1\right)^2-1\right]}\)

    =\(\frac{-1.\left(2^2-1\right)...\left(20^2-1\right)\left(22^2-1\right)}{\left(2^2-1\right)\left(4^2-1\right)...\left(22^2-1\right)\left(24^2-1\right)}\)

     =-1

                                              kết bạn với mk nha !!!!!^-^