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\(\frac{27\cdot18+27\cdot103-120\cdot27}{15\cdot33+33\cdot12}\)
\(=\frac{27\left(18+103-120\right)}{33\cdot\left(15+12\right)}\)
\(=\frac{27}{33\cdot3}=\frac{27}{99}\)
\(=\frac{3}{11}\)
Đề bài : Tính
\(\frac{27.18+27.103-120.27}{15.33+33.12}\)
\(=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)
\(=\frac{27}{33.27}\)
\(=\frac{1}{33}.\)
\(=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)
\(=\frac{27\times1}{33\times27}\)
\(=\frac{1}{33}\)
Kết quả của phép tính:
(-2) + (-18)= -20
(-5) + 33= 28
(-35) + 24= -11
khẳng định đúng : (- 25 )> (-27)
1: =66+34-73-27+99=99
2: =54-45+30-54-30+5=-40
3: =42-56+33-33+56-42=0
4: =50-47-50+18+47-18=0
5: =10-28+30-4=2-4+10=12-4=8
\(\text{1.-(-66)+34+(-73)-27+99}.\)
\(=66+34-73-27+99=100-100+99=-99.\)
\(\text{2.54-(45-30+54)-30+5}.\)
\(54-45+30-54-30+5=\left(54-54\right)+\left(-45+5\right)+\left(30-30\right)=-40.\)
\(\text{3.42-56+33-(33-56+42)}.\)
\(=42-56+33-33+56-42=\left(42-42\right)+\left(-56+56\right)+\left(33-33\right)=0.\)
\(\text{4.50-(47+50-18)+(47-18)}.\)
\(=50-47-50+18+47-18=\left(50-50\right)+\left(-47+47\right)+\left(18-18\right)=0.\)
\(\text{5.10-(28-30+4)}.\)
\(=10-28+30-4=10+30-20-8-4=20-12=8.\)
\(\text{6.-(-73)+(44-94+27)+94}.\)
\(=73+44-94+27+94=\left(73+27\right)+\left(-94+94\right)+44=100+44=144.\)
\(\frac{27.18+27.103-120.27}{15.33+33.12}\)
\(=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)
\(=\frac{27.1}{33.27}\)
\(=\frac{1}{33}\)
\(\frac{1}{33}\)