Bài 109 trang 49 SGK Toán 6 tập 2

Tính bằng hai cách:

a) \( \displaystyle 2{4 \over 9} + 1{1 \over 6}\)

b) \( \displaystyle 7{1 \over 8} - 5{3 \over 4}\)

c) \( \displaystyle 4 - 2{6 \over 7}\) 

Lời giải

Cách 1. 

a) \( \displaystyle 2{4 \over 9} + 1{1 \over 6} = {{22} \over 9} + {7 \over 6} = {{44 + 21} \over {18}}\)\( \displaystyle = {{65} \over {18}} = 3{{11} \over {18}}\)

b) \( \displaystyle 7{1 \over 8} - 5{3 \over 4} = {{57} \over 8} - {{23} \over 4} = {{57 - 46} \over 8} \)\( \displaystyle = {{11} \over 8} = 1{3 \over 8}\)

c) \( \displaystyle 4 - 2{6 \over 7} = {{28} \over 7} - {{20} \over 7} \)\( \displaystyle = {8 \over 7} = 1{1 \over 7}\)

Cách 2.

a) \( \displaystyle 2{4 \over 9} + 1{1 \over 6} = \left( {2 + 1} \right) + \left( {{4 \over 9} + {1 \over 6}} \right) \)\( \displaystyle = 3 + {{8 + 3} \over {18}} = 3{{11} \over {18}}\)

b) \( \displaystyle 7{1 \over 8} - 5{3 \over 4} = \left( {7 - 5} \right) + \left( {{1 \over 8} - {3 \over 4}} \right) \)\( \displaystyle = 2 - {5 \over 8} = {{11} \over 8} = 1{3 \over 8}\)

c) \( \displaystyle 4 - 2{6 \over 7} = \left( {4 - 2} \right) - {6 \over 7} \)\( \displaystyle = 2 - {6 \over 7} = {8 \over 7} = 1{1 \over 7}\)  


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