Bài 1: Tính:
a) \(\left( { - 2{1 \over 5}} \right).\left( {{{ - 9} \over {11}}} \right)\left( { - 1{1 \over {14}}} \right).{2 \over 5}\)
b) \( - 6\left( {{{ - 2} \over 3}} \right).0,25\)
Bài 2: Tính:
a) \(1 + {1 \over {1 + {1 \over 2}}}\)
b) \(1 + {1 \over {1 + {1 \over {1 + {1 \over 2}}}}}\)
Bài 1:
a) \(\left( { - 2{1 \over 5}} \right).\left( {{{ - 9} \over {11}}} \right)\left( { - 1{1 \over {14}}} \right).{2 \over 5} \)
\(= \left( { - {{11} \over 5}} \right).\left( {{{ - 9} \over {11}}} \right).\left( { - {{15} \over {14}}} \right).{2 \over 5}\)
\(={{\left( { - 11} \right).\left( { - 9} \right).\left( { - 15} \right).2} \over {5.11.14.5}} = {{ - 11.9.15.2} \over {5.11.14.5}} = {{ - 27} \over {35}}.\)
b) \( - 6\left( {{{ - 2} \over 3}} \right).0,25 = - 6\left( { - {2 \over 3}} \right).{1 \over 4} = {{6.2.1} \over {3.4}}\)\(\, = 1\)
Bài 2:
a) \(1 + {1 \over {1 + {1 \over 2}}} = 1 + {1 \over {{3 \over 2}}} = 1 + {2 \over 3} = {5 \over 3}\)
b) \(1 + {1 \over {1 + {1 \over {1 + {1 \over 2}}}}} = 1 + {1 \over {1 + {1 \over {{3 \over 2}}}}} = 1 + {1 \over {1 + {2 \over 3}}} \)\(\;= 1 + {1 \over {{5 \over 3}}} = 1 + {3 \over 5} = {8 \over 5}.\)
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