Giải bài 28 trang 37 SBT Toán 6 Cánh diều tập 2 - CD

Hướng dẫn giải

Tính A và B rồi so sánh

Lời giải chi tiết

a) \(A = \frac{1}{2} + \frac{{ - 3}}{8} + \frac{5}{9} = \frac{4}{8} + \frac{{ - 3}}{8} + \frac{5}{9} = \frac{1}{8} + \frac{5}{9} = \frac{{1.9 + 5.8}}{{8.9}} = \frac{{49}}{{72}}\)

\(B = \frac{{13}}{{ - 30}} + \frac{{17}}{{45}} + \frac{{ - 7}}{{18}} = \frac{{ - 39}}{{90}} + \frac{{34}}{{90}} + \frac{{ - 35}}{{90}} = \frac{{ - 39 + 34 + ( - 35)}}{{90}} = \frac{{ - 40}}{{90}} = \frac{{ - 4}}{9}\)

Vì \(\frac{{ - 4}}{9} < 0 < \frac{{49}}{{72}}\) nên B < A.

b) \(C = \frac{{12}}{{25}} + \frac{{ - 8}}{{15}} + \frac{{ - 4}}{9} = \frac{{108}}{{225}} + \frac{{ - 120}}{{225}} + \frac{{ - 100}}{{225}} = \frac{{108 + ( - 120) + ( - 100)}}{{225}} = \frac{{ - 112}}{{225}}\)

\(D = \frac{{ - 5}}{{12}} + \frac{4}{9} + \frac{{11}}{{ - 6}} = \frac{{ - 15}}{{36}} + \frac{{16}}{{36}} + \frac{{ - 66}}{{36}} = \frac{{ - 15 + 16 + ( - 66)}}{{36}} = \frac{{ - 65}}{{36}}\)

Mà C > -1 > D nên C > D.

c) \(M = \frac{1}{3} + \frac{2}{{ - 5}} + \frac{7}{2} = \frac{{10}}{{30}} + \frac{{ - 12}}{{30}} + \frac{{105}}{{30}} = \frac{{10 + ( - 12) + (105)}}{{30}} = \frac{{103}}{{30}}\)

\(\begin{array}{l}N = \frac{{19}}{{ - 7}} + \frac{{21}}{5} + \frac{{ - 2}}{7} = \left( {\frac{{19}}{{ - 7}} + \frac{{ - 2}}{7}} \right) + \frac{{21}}{5} = \left( {\frac{{ - 19}}{7} + \frac{{ - 2}}{7}} \right) + \frac{{21}}{5}\\ = \frac{{ - 21}}{7} + \frac{{21}}{5} =  - 3 + \frac{{21}}{5} = \frac{{ - 15 + 21}}{5} = \frac{6}{5}\end{array}\)

Mà \(\frac{6}{5} = \frac{{36}}{{30}} < \frac{{103}}{{30}}\)

Vậy N < M.

d) \(P = \frac{{34}}{{24}} + \frac{{ - 8}}{{15}} + \frac{1}{{10}} = \frac{{17}}{{12}} + \frac{{ - 8}}{{15}} + \frac{1}{{10}} = \frac{{85}}{{60}} + \frac{{ - 32}}{{60}} + \frac{6}{{60}} = \frac{{59}}{{60}}\)

\(Q = \frac{8}{{21}} + 1 + \frac{1}{{ - 21}} = \left( {\frac{8}{{21}} + \frac{1}{{ - 21}}} \right) + 1 = \left( {\frac{8}{{21}} + \frac{{ - 1}}{{21}}} \right) + 1 = \frac{7}{{21}} + 1 = \frac{1}{3} + 1 = \frac{4}{3}\)

Mà \(\frac{4}{3} > 1 > \frac{{59}}{{60}}\)

Vậy P < Q

-- Mod Toán 6 HỌC247