Bài 1: Ta chỉ cần bỏ ngoặc rồi cộng hai phân số để ra kết quả là số tự nhiên là xong

Bài 2:

A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+............+\frac{1}{2003.2004}\)

A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.............-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)

A = \(1-\frac{1}{2004}\)

A = \(\frac{2004}{2004}-\frac{1}{2004}=\frac{2003}{2004}\)

Mk làm mẫu 1 bài nha 

Bài 1 :

a, = (1/4+3/4) - (5/13+8/13)+2/11

    = 1 - 1 + 2/11

    = 2/11

Tk mk nha

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(\Leftrightarrow A=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\right)\)

\(\Leftrightarrow A=1\left(1-\frac{1}{2004}\right)\)

\(\Leftrightarrow A=\frac{2003}{2004}\)

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

đúng rùi

\(\left(\frac{3}{4}+\frac{2}{11}-\frac{8}{13}\right)+\left(\frac{1}{4}-\frac{5}{13}\right)\)

\(=\frac{3}{4}+\frac{2}{11}-\frac{8}{13}+\frac{1}{4}-\frac{5}{13}\)

\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{2}{11}-\left(\frac{8}{13}+\frac{5}{13}\right)\)

\(=1+\frac{2}{11}-1\)

\(=\frac{2}{11}\)

\(\left(\frac{9}{53}+\frac{21}{31}\right)+\left(\frac{44}{53}-\frac{16}{7}\right)+\frac{10}{31}\)

\(=\frac{9}{53}+\frac{21}{31}+\frac{44}{53}-\frac{16}{7}+\frac{10}{31}\)

\(=\left(\frac{9}{53}+\frac{44}{53}\right)+\left(\frac{21}{31}+\frac{10}{31}\right)-\frac{16}{7}\)

\(=1+1-\frac{16}{7}\)

\(=2-\frac{16}{7}\)

\(=\frac{14}{7}-\frac{16}{7}\)

\(=\frac{-2}{7}\)

a) \(\dfrac{3}{4}=\dfrac{3x4}{4x4}=\dfrac{12}{16},\dfrac{6}{7}=\dfrac{6x2}{7x2}=\dfrac{12}{14}\)

Do 16 > 14 => \(\dfrac{12}{16}< \dfrac{12}{14}hay\dfrac{3}{4}< \dfrac{6}{7}\)

1216<1214hay34<67

a,  -3/4 + 3/7 + -1/4 + 4/9 + 4/7

=\(\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)

=\(-1+1+\frac{4}{9}\)

=\(\frac{4}{9}\)

b, -7/9 x 4/11 + -7/9 x 7/11 + 5 và 7/9 

\(=\frac{7}{9}x\frac{-4}{11}+\frac{7}{9}x\frac{-7}{11}+5x\frac{7}{9}\)

\(=\frac{7}{9}x\left(\frac{-4}{11}+\frac{-7}{11}+5\right)\)

\(=\frac{7}{9}x4\)

\(=\frac{28}{9}\)

c, ( 21/31 + 2013/6039 ) - ( 44/53 - 10/31) - /53

a 16,22

b31,43

c121,169

d2223,222223

e25,35

tí nữa lm nốt cho

a. 16;22

b. 31;43