Bài 1 trang 88 SGK Toán 7 tập 2

Đề bài

Thực hiện các phép tính:

a) \(9,6.2\dfrac{1}{2} - \left( {2.125 - 1\dfrac{5}{{12}}} \right):\dfrac{1}{4}\)

b) \(\dfrac{5}{{18}} - 1,456:\dfrac{7}{{25}} + 4,5.\dfrac{4}{5}\);

c) \(\left( {\dfrac{1}{2} + 0,8 - 1\dfrac{1}{3}} \right).\left( {2,3 + 4\dfrac{7}{{25}} - 1,28} \right)\)

d) \(\left( { - 5} \right).12:\left[ {\left( { - \dfrac{1}{4}} \right) + \dfrac{1}{2}:\left( { - 2} \right)} \right] + 1\dfrac{1}{3}\).

Lời giải

a) 

\(\eqalign{
& 9,6.2{1 \over 2} - \left( {2.125 - 1{5 \over {12}}} \right):{1 \over 4} \cr
& = 9,6.{5 \over 2} - \left( {250 - {{17} \over {12}}} \right) \times 4 \cr
& = 4,8.5 - \left( {1000 - {{17} \over 3}} \right) \cr
& = 24 - 1000 + {{17} \over 3} \cr
& = - 976 + {{17} \over 3} \cr
& = - 976 + {{18 - 1} \over 3} \cr
& = - 976 + 6 - {1 \over 3} \cr
& = - 970{1 \over 3} \cr} \)

b)

\(\eqalign{
& {5 \over {18}} - 1,456:{7 \over {25}} + 4,5.{4 \over 5} \cr
& = {5 \over {18}} - 1,456 \times {{25} \over 7} + {9 \over 2}.{4 \over 5} \cr
& = {5 \over {18}} - 0,208 \times 25 + {{18} \over 5} \cr
& = {5 \over {18}} - 5,2 + {{18} \over 5} \cr
& = {{25 - 468 + 324} \over {90}} \cr
& = {{ - 119} \over {90}} \cr} \)

c) 

\(\eqalign{
& \left( {{1 \over 2} + 0,8 - 1{1 \over 3}} \right).\left( {2,3 + 4{7 \over {25}} - 1,28} \right) \cr
& = \left( {{1 \over 2} + {4 \over 5} - {4 \over 3}} \right).\left( {{{23} \over {10}} + {{107} \over {25}} - {{32} \over {25}}} \right) \cr
& = \left( {{{15 + 24 - 40} \over {30}}} \right).\left( {{{23} \over {10}} + {{107} \over {25}} - {{32} \over {25}}} \right) \cr
& = \left( {{{15 + 24 - 40} \over {30}}} \right).\left( {{{115 + 214 - 64} \over {50}}} \right) \cr
& = {{ - 1} \over {30}}.{{265} \over {50}} = {{ - 53} \over {300}} \cr} \)

d) 

\(\eqalign{
& \left( { - 5} \right).12:\left[ {\left( { - {1 \over 4}} \right) + {1 \over 2}:\left( { - 2} \right)} \right] + 1{1 \over 3} \cr
& = - 60:\left[ { - {1 \over 4} + {1 \over 2} \times \left( {{{ - 1} \over 2}} \right)} \right] + 1{1 \over 3} \cr
& = - 60:\left[ { - {1 \over 4} - {1 \over 4}} \right] + 1{1 \over 3} \cr
& = - 60:\left( { - {1 \over 2}} \right) + 1{1 \over 3} \cr
& = - 60.( - 2) + 1{1 \over 3} \cr
& = 120 + 1{1 \over 3} \cr
& = 121{1 \over 3} \cr} \)