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a) 998.34=(1000−2).34=34000−68=33932
b,\(1990.1990-\left(1990+2\right).1988=1990.1990-1990.1988-2.1988=1990\left(1990-1988\right)-2.1988=1990.2-2.1998=2\left(1990-1998\right)=2.2\)= 4
c. \(\text{(1374.57+678.86).(26.13+74.14)}\)
Câu này đề hơi lạ
a, 1990 . 1990 - 1992 . 1988
= 3960100 - 3960096
= 4 .
b, ( 1374 . 57 + 687 . 86 ) : ( 26 . 13 + 74 . 14 )
= ( 78318 + 59082 ) : ( 338 + 1036 )
= 137400 : 1374
= 100 .
c, \(\frac{37.13-13}{24 +37.12}\)
\(=\frac{37.\left(12+1\right)-13}{24+37.12}\)
\(=\frac{37.12+37-13}{24+37.12}\)
\(=\frac{37.12+24}{24+37.12}\)
\(=1\)
a. 19991999.1998 - 19981998.1999
=1999.10001.1998 - 1998.10001.1999
=0
b.(1374.57+687.86):(26.13+74.14)
=(687.2.57+687.86):(338+1036)
=\([687.\left(2.57+86\right)]:1374\)
=678.200:1374
=137400:1374
=100
A= 19 . 64+ 76.34
A=19.2.32+38.2.34
A=38.32+38.68
A=38.(32+68)
A=38.100
A=3800
B= 35 .12+ 65.13
B= 35.12+65.(12+1)
B=35.12 +65.12+65
B=(35+65).12 +65
B=100.12+65
B=1200+65
B=1265
C=136.68+16.272
C=136.68+16.2.136
C=136.68+32.136
C=136.(68+32)
C=136.100
C=13600
D=(2+4+6+...+100).(36.333-108.111)
D=(2+4+6+...+100).(36.333-36.3.111)
D=(2+4+6+...+100).(36.333+36.333)
D=(2+4+6+..+100).0
D=0
E=19991999.1998+19981998.1999
E=1999.10001.1998-1998.10001.1999
E=0
A= 19.64 +76.34 B=35.12 +65.13
=76(16+34) =35.13-35+65.13
=76.50 =1300-35
=380 =1765
a)
1990.1990 -1992.1998 = (1995-5)^2 -(1995+3)(1995-3) = -10.1995 +25 +9 = -(19950-34) = -19916
b)
ta có
tu = 124 .237 +152 = 122.137 + 474 +152= 122. 137 + 626
mau = ( 870 + (237-2).122= 870 +122.237 - 244 =122.237 + 626
=> phân thức đã cho = tử/mẫu = 1
a/ \(=1990.1990-\left(1990+2\right).\left(1990-2\right)=1990.1990-\left(1990.1990-4\right)=4\)
b/ \(=\frac{2.687.57+687.86}{26.13+74.\left(13+1\right)}=\frac{687\left(2.57+86\right)}{26.13+74.13+74}=\frac{687.\left(114+86\right)}{13\left(26+74\right)+74}=\)
\(=\frac{687.200}{13.100+74}=\frac{687.200}{1374}=\frac{687.200}{687.2}=\frac{200}{2}=100\)
c/ \(=\frac{214.237+152}{870+\left(237-2\right).\left(124-2\right)}=\frac{124.237+152}{870+124.237-2.237-2.124+4}=\)
\(=\frac{124.237+152}{124.237+870+4-2.237-2.124}=\frac{124.237+152}{124.237+152}=1\)
a)1990*1990-1992*1988=1990*(1988+2)-(1990+2)*1988=1990*1988+2*1990-1990*1988-2*1988=(1990*1988-1990*1988)+(2*1990-2*1988)=0+2*(1990-1988)=2*2=4
b)(1374*57+687*86)/(26*13+74*14)=(687*2*57+687*86)/(26*13+74+74*13)=(687*114+687*86)/(100*13+74)
=(687*200)/1374=137400/1374=100
c)(124*237+152)/(870+235*122)=\(\frac{\left(122+2\right)\cdot237+152}{870+\left(237-2\right)\cdot122}=\frac{122\cdot237+2\cdot237+152}{870-2\cdot122+237\cdot122}=\frac{122\cdot237+626}{626+237\cdot122}=1\)
d)\(\frac{423134\cdot846267-423133}{846267\cdot423133+423134}=\frac{\left(423133+1\right)\cdot846267-423133}{846267\cdot423133+423134}\)
\(=\frac{423133\cdot846267+846267-423133}{846267\cdot423133+423134}=\frac{423133\cdot846267+423134}{846267\cdot423133+423134}=1\)
(mỏi hết cả mắt)
a) 19.64+76.34=19.64+19.4.34=19.(64+136)=19.200=3800
b) 35.12+65.13=35.12+65.12+65=12.(35+65)+65=12.100+65=1265
c) 136.68+16.272=136.68+32.136=136.(68+32)=136.100=13600
d) ( 2+4+6+...+100).(36.333 - 108.111)=(....).(108.111 - 108.111)=(....).0=0
a) 19. 64+76.34
=19 . 2. 32+ 38.2 . 34
=38.32+38.68
=38. (32+68)
=38.100
=3 800
b)35.12+65.13
=35.12+65.(12+1)
=35.12+65.12+65
=12.(35+65)+65
=12.100+65
=1 200 +65
=1265
c1:
=19.64+19.4.34
=19.(64+4.34)
=19.200
=3800
c2:
=35.12+65.12+65.1
=12.(35+65)+65
=12.100+65
=1200+65
=1265
c3
=136.68+16.136.2
=136.(68+16.2)
=136.100
=13600
c4:ta sẽ giải vế phía sau trước
=>36.3.111-108.111
=>108.111-108.111
=>0
KL: kết quả phép tính =0 vì (bn lớp 6 rồi lên chắc hiểu và tự giải thích đc đúng ko)
c5:
=1999.(1999.1998-1998.1999)
=1999.[1998.(1999-1998)]
=1999.[1998.1]
=1999.1998
=(200-1).1998
=200.1998-1.1998
=399600-1998
=397602
cho mk cái ****
a) Xét 1992.1998 = (1995 - 3).(1995 + 3) = 1995.(1995 + 3) - 3.(1995 + 3) = 19952 + 1995.3 - 3.1995 - 32 = 19952 - 9
Vậy 19952 - 1992.1998 = 19952 - (19952 - 9) = 9