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198 : [130 – (27 – 19 )^2] + 2021^0
= 198: [130 - 8^2] + 1
= 198 :( 130- 64) +1
= 198: (66) +1
= 3+1=4
a) 25.69+31.25-25.6
= 25.(69+31-6)
= 25.94
= 2350
b) 198:[130-(27-19)2]+20210
= 198:[130-82]+1
= 198:(130-64)+1
= 198:66+1
= 3+1
= 4
c) 520:[515.(15+10)]
= 520:[515.25]
= 520:(515.52)
= 520: 517
= 53
a: =25x100-150=2500-150=2350
c: \(=520:\left\{515\cdot25\right\}\)
=104/2575
a) Ta có: \(\left(-68+42\right)-\left(5042-6068\right)-\left(-2\right)^0\)
\(=-68+42-5042+6068-1\)
\(=6000-5000-1\)
\(=999\)
b) Ta có: \(29\cdot\left(19-37\right)-19\cdot\left(29-37\right)\)
\(=29\cdot19-29\cdot37-29\cdot19+19\cdot37\)
\(=-29\cdot37+19\cdot37\)
\(=37\cdot\left(-29+19\right)\)
\(=37\cdot\left(-10\right)=-370\)
c) Ta có: \(\left(-15\right)\cdot24+15\cdot\left(-75\right)-15\)
\(=\left(-15\right)\cdot24+\left(-15\right)\cdot75+\left(-15\right)\cdot1\)
\(=\left(-15\right)\cdot\left(24+75+1\right)\)
\(=-15\cdot100=-1500\)
d) Ta có: \(\frac{1}{5}+\frac{4}{7}-\frac{11}{35}\)
\(=\frac{5}{35}+\frac{20}{35}-\frac{11}{35}\)
\(=\frac{14}{35}=\frac{2}{5}\)
e) Ta có: \(\left(13\cdot95-73\right)-\left(13\cdot45+27\right)-\left(-1\right)^{2021}\)
\(=13\cdot95-73-13\cdot45-27-\left(-1\right)\)
\(=13\left(95-45\right)-\left(73+27\right)+1\)
\(=13\cdot50-100+1\)
\(=551\)
Bài 1: Tính hợp lý
a) ( - 68 + 42 ) - ( 5042 - 6068 ) - (-2)\(^0\)
= -26 - (-1026) - 1
= - 26 + 1026 - 1
= 1000 - 1
= 999
Mấy câu còn lại c hơi làm biếng, sr :<<
1)tính
a)\(-18+\left(-19\right)+27\)
\(=-18-19+27\)
\(=-\left(18+19\right)+27\)
\(=-37+27\)
\(=-\left(37-27\right)\)
\(=-10\)
b)\(-198-194+286\)
\(=-\left(198+194\right)+286\)
\(=-392+286\)
\(=-\left(392-286\right)\)
\(=-6\)
c)\(-19-\left(-17\right)\)
\(=-19+17\)
\(=-\left(19-17\right)\)
\(=-2\)
d)\(-48+\left(-47\right)-\left(-15\right)\)
\(=-48-47+15\)
\(=-\left(48+47\right)+15\)
\(=-95+15\)
\(=-\left(95-15\right)\)
\(=-80\)
e)\(23-\left(-16\right)+\left(-80\right)\)
\(=23+16-80\)
\(=39-80\)
\(=-80+39\)
\(=-\left(80-39\right)\)
\(=-41\)
f)\(99-\left(-99\right)-198\)
\(=99+99-198\)
\(=198-198\)
\(=0\)
4
=4