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\(\frac{3}{13}+4\frac{7}{26}+1\frac{1}{12}=5+\left(\frac{3}{13}+\frac{7}{26}+\frac{1}{12}\right)=5\frac{7}{12}\)
\(\frac{2}{3}+\frac{4}{37}+5\frac{5}{11}=5+\left(\frac{2}{3}+\frac{4}{27}+\frac{5}{111}\right)\)ta thấy 2/3 > 7/12
Nên \(\frac{3}{13}+4\frac{7}{26}+1\frac{1}{2}\)< \(\frac{2}{3}+\frac{4}{37}+5\frac{5}{111}\)
\(\left(\dfrac{5}{7}-\dfrac{7}{7}\right)-\left[0,2-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)
=\(-\dfrac{2}{7}-\left[\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
=\(-\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{10}\)
=\(\left(-\dfrac{2}{7}-\dfrac{2}{7}\right)-\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\left(\dfrac{2}{10}+\dfrac{1}{10}\right)\)
=\(-\dfrac{4}{7}-\dfrac{3}{10}\)
=\(-\dfrac{40}{70}-\dfrac{21}{70}\)
=\(-\dfrac{61}{70}\)
(3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\)) - (5 - \(\dfrac{1}{3}\) - \(\dfrac{5}{6}\)) - (6 - \(\dfrac{7}{4}\) - \(\dfrac{3}{2}\))
= 3 - \(\dfrac{1}{4}\) + \(\dfrac{2}{3}\) - 5 + \(\dfrac{1}{3}\) + \(\dfrac{5}{6}\) - 6 + \(\dfrac{7}{4}\) + \(\dfrac{3}{2}\)
= (3 - 5 - 6) + ( \(\dfrac{7}{4}\) - \(\dfrac{1}{4}\)) + (\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= - 8 + \(\dfrac{3}{2}\) + 1 + \(\dfrac{3}{2}\) + \(\dfrac{5}{6}\)
= (- 8 + 1) + (\(\dfrac{3}{2}\) + \(\dfrac{3}{2}\)) + \(\dfrac{5}{6}\)
= -7 + 3 + \(\dfrac{5}{6}\)
= - 4 + \(\dfrac{5}{6}\)
= \(\dfrac{-19}{6}\)
Bài 1:
a: -8/12<0<-3/-4
b: -56/24<0<7/3
c: 4/25<1<15/13
=>-4/25>-15/13
Bài 2:
a: =-60/45=-4/3
b: =4/15-3/2-8/5=8/30-45/30-48/30=-85/30=-17/6

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{13.14}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=\dfrac{1}{1}-\dfrac{1}{14}=\dfrac{13}{14}\)
Ta thấy: \(\dfrac{13}{14}< 1\)
Vậy A<1
thank bạn nhiều
Bước cuối
\(1-\dfrac{1}{14}< 1\) vì đang so sánh với 1 cũng là lý do đề bắt so sánh với 1
xét mặt toán học hay hơn cộng vào thành 13/14 < 1
nếu đề bắt so sánh với a/b thì khi đó mới phải mất công quy đồng lên
ý kiến chủ quan theo mình là vậy
Cám ơn ạ
Đặt \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{13\cdot14}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=1-\dfrac{1}{14}\)
\(=\dfrac{14}{14}-\dfrac{1}{14}\)
\(=\dfrac{13}{14}< 1\)
Vậy A < 1
Ta có :B= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{13.14}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+....+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=1-\dfrac{1}{14}\)
Vì \(1-\dfrac{1}{14}< 1\)
Suy ra : \(B< 1\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{13\cdot14}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=1-\dfrac{1}{14}=\dfrac{13}{14}< 1\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+....+\dfrac{1}{13\cdot14}< 1\)
Vậy \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{13\cdot14}< 1\)