-126-(4^2-5)^2+|-870|:29
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a: \(=625-44\cdot12+\dfrac{41}{17}=97+\dfrac{41}{17}=\dfrac{1690}{17}\)
b: \(=-126-11^2+30=-126-121+30=-217\)
\(A=625-\left(61-17\right)\cdot12+\left(27+24\right)\div17\)
\(A=625-44\cdot12+51\div17\)
\(A=625-528+3\)
\(A=100\)
\(B=-126-\left(4^2-5^2\right)+870-29\)
\(B=-126-\left(16-25\right)+870-29\)
\(B=-126-\left(-9\right)+870-29\)
\(B=-117+870-29\)
\(B=753-29\)
\(B=724\)
Chúc bn hok tốt
1) A \(=625-\left(61-17\right).12+\left(27+24\right):17\)
\(\Leftrightarrow A=625-44.12+51:17\)
\(\Leftrightarrow A=625-528+3\)
\(\Leftrightarrow A=100\)
Vậy A = 100
2) B \(=-126-\left(4^2-5^2\right)+870-29\)
\(\Leftrightarrow B=-126-\left(16-25\right)+870-29\)
\(\Leftrightarrow B=-126-16+25+870-29\)
\(\Leftrightarrow B=724\)
Vậy B = 724
a,\(\left(x-3\right).\left(2y+1\right)=7\)
Vì \(x;y\inℤ=>x-3;2y+1\inℤ\)
\(=>x-3;2y+1\inƯ\left(7\right)\)
Nên ta có bảng sau
x-3 | 1 | 7 | -7 | -1 |
2y+1 | 7 | 1 | -1 | -7 |
x | 4 | 10 | -4 | 2 |
y | 3 | 0 | -1 | -4 |
Vậy ...
b,\(A=-126-\left(4^2-5\right)^2+870:29\)
\(=-126-\left(16-5\right)^2+30\)
\(=-126-11^2+30\)
\(=-247+30=-217\)
b) 4.(–5)2 + 2.(–5) – 20
= 4.(-5)2 + 2.(-5) - (-4)(-5)
= (-5).[4.(-5) + 2 - (-4)]
= (-5).[-20 + 2 + 4]
=(-5).(-14)
=70
a) A=625–(61–17).12+(27+27):17A=625–(61–17).12+(27+27):17
=625–528+3=100=625–528+3=100
b) B=−126–(42–5)2+870:29B=−126–(42–5)2+870:29
=−126–112+30=−126–112+30
=−126–121+30=−217
a)C= (-124)+ (36 + 124 - 99 ) - ( 136 - 1 ) = (-124) +36 +124 -99 - 136 +1
= -198
b) D = { 115+[ 32 - ( 132 -5 )] } +(-25) +(-25)
= {115+[32-132+5]} +(-25) + (-25)
= {115+(-95)} + (-25)+ (-25)
= 20 +(-25) +(-25) = -30
c)F = [(123 - 17 ) - (123 + 33 ) ] - { 34 - [ 34 + ( 57 -50 ) -7 ] }
= [123 -17 -123 -33 ] - {34- [34 + 57-50 -7]}
= -50 - {34-34} = -50 - 34 +34 = -50
d) mik bo nha
e) E = (-7) +( -2020) -(-7) +2020
= [(-7)-(-7)]+[(-2020) +2020]= 0
g)B= (-2019)-( 29 - 2019) = (-2019) - 29 +2019 = -29
co moi cau d mik ko biet lam :v