Tính: d) ( 1/25 − 0 , 4 )^2 : 9/125 − [ ( 1(1/3) − 2/5 ) . 3/7 ]
Giải thích
d) (125−0,4)2:9125−[(113−25).37]
=(125−25)2.1259−[(43−25).37]
=(−925)2.1259−[1415.37]
=92252.1259−25=(32)2(52)2.5332−25
=3454.5332−25=325−25
=95−25=75
d) (125−0,4)2:9125−[(113−25).37]
=(125−25)2.1259−[(43−25).37]
=(−925)2.1259−[1415.37]
=92252.1259−25=(32)2(52)2.5332−25
=3454.5332−25=325−25
=95−25=75