a, \(=\dfrac{1+4}{5}+\dfrac{5+1+3}{9}=1+1=2\)

b, \(=\dfrac{1+4+2}{3}+\dfrac{1+2+5}{6}=\dfrac{6}{3}+\dfrac{8}{6}=2+\dfrac{4}{3}=\dfrac{6+4}{3}=\dfrac{10}{3}\)

cảm ơn bạn nha

Cách 1 :

Ta có :

\(A=\dfrac{10^5+4}{10^5-1}=\dfrac{10^5-1+4+1}{10^5-1}=\dfrac{10^5-1+5}{10^5-1}=1+\dfrac{5}{10^5-1}\)

\(B=\dfrac{10^5+3}{10^5-2}=\dfrac{10^5-2+3+2}{10^5-2}=\dfrac{10^5-2+5}{10^5-2}=1+\dfrac{5}{10^5-2}\)

Vì \(1+\dfrac{5}{10^5-1}< 1+\dfrac{5}{10^5-2}\Rightarrow A< B\)

Cách 2 chưa nghĩ ra!!

~ Học tốt ~

C2: Dễ thấy \(\dfrac{10^5+3}{10^5-2}>1\)

\(\Rightarrow B=\dfrac{10^5+3}{10^5-2}>\dfrac{10^5+3+1}{10^5-2+1}=\dfrac{10^5+4}{10^5-1}=A\)

Vậy \(B>A\)

Bài 3

a,26/100+0,009+41/100+0,24

0,26+0,09+0,41+0,24

(0,26+0,24)+(0,09+0,41)

0,5+0,5

=1

b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4

(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)

30+1+1+1+1

=34

Bài 4,5 khó quá mik ko bít lamf^^))

Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1

           \(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)

         b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)

Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)

a)

C1: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)

\(=\dfrac{9}{4}+\dfrac{7}{6}\)

\(=\dfrac{27}{12}+\dfrac{14}{12}\)

\(=\dfrac{41}{12}=3\dfrac{5}{12}\)

C2: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)

\(=2\dfrac{3}{12}+1\dfrac{2}{12}\)

\(=3\dfrac{5}{12}\)

Đáp án và hướng dẫn giải bài 109:

a) Cách 1

cách 2

b) Cách 1

cách 2

c, cách 1

cách 2

a)=2

b)=1/3

\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)

\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)

-_-

1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)

\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)

\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)

\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)

=1

\(=\dfrac{17}{2}+\dfrac{3}{5}\times\left(1+\dfrac{1}{2}+1\right)=\dfrac{17}{2}+\dfrac{3}{5}\times\dfrac{5}{2}=\dfrac{17}{2}+\dfrac{15}{10}=\dfrac{17}{2}+\dfrac{3}{2}=\dfrac{17+3}{2}=\dfrac{20}{2}=10\)

\(\dfrac{17}{2}+\dfrac{3}{5}+\dfrac{3}{5}\times\dfrac{1}{2}+\dfrac{3}{5}=\dfrac{17}{2}+\dfrac{3}{5}+\dfrac{3}{10}+\dfrac{3}{5}=\dfrac{85}{10}+\dfrac{6}{10}+\dfrac{3}{10}+\dfrac{6}{10}=10\)