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\(\dfrac{47}{53}\left(\dfrac{17}{3}-\dfrac{53}{47}\right)+\dfrac{17}{3}\left(\dfrac{6}{17}-\dfrac{47}{53}\right)\)
\(=\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{45}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}-\dfrac{17}{3}.\dfrac{47}{53}\)
\(=-1+\dfrac{6}{3}=-1+2=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-53\right)\cdot49+\left(-49\right)\cdot47\)
\(=49\left(-53-47\right)\)
\(=49\cdot\left(-100\right)\)
\(=-4900\)
(−53)⋅49+(−49)⋅47
(−53)⋅49+(−49)⋅4
=49(−53−47)=49(−53−47)
=49⋅(−100)=49⋅(−100)
=−4900
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Bài 1:
a) - | 12 - 14 | - | 98 - 15 + 2 | - 14
= - | - 2 | - | 85 | - 14
= -2 - 85 - 14
= -101
b) ( -24 ) - ( 12 - 19 ) - | 15 - 12 - 7 |
= ( -24 ) - ( - 7 ) - | -4 |
= ( -24 ) + 7 - 4
= -21
Bài 2:
a) 5 - ( 1997 - 2005 ) + 1997
= 5 - 1997 + 2005 + 1997
= ( 5 + 2005 ) + ( 1997 - 1997 )
= 2010 + 0
= 2010
b) 1579 - ( 53 + 1679 ) - ( -53 )
= 1579 - 53 - 1679 + 53
= ( 1579 - 1679 ) + ( 53 - 53 )
= ( -100 ) + 0
= -100
c) -48725 - ( 357 - 48725 ) + 300
= -48725 - 357 + 48725 + 300
= [( -48725 ) + 48725 ] + ( 357 - 300 )
= 0 + ( -57 )
= -57
d) 4567 + ( 1234 - 4567 ) - 4
= 4567 + 1234 - 4567 - 4
= ( 4567 - 4567 ) + ( 1234 - 4 )
= 0 + 1230
= 1230
a) \(-\left|12-14\right|-\left|98-15+2\right|-14\)
\(=-\left|2\right|-\left|85\right|-14\)
\(=\left(-2\right)-85-14\)
\(=\left(-2\right)-\left(85+14\right)\)
\(=\left(-2\right)-99\)
\(=\left(-2\right)+\left(-99\right)=-101\)
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( 235 - 47 ) + ( 175 - 53 )
= 235 - 47 + 175 - 53
= [ 235 - 175 ] + [ -47 - ( -53 ) ]
= 60 + 6
= 66
Vậy kết quả là 66
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`47/53*(17/3-53/47)+17/3*(6/17-47/53)`
`=47/53*17/3-1+6/3-17/3*47/53`
`=2-1+17/3*47/53-17/3*47/53`
`=1`
\(\dfrac{47}{53}.\left(\dfrac{17}{3}-\dfrac{53}{47}\right)+\dfrac{17}{3}.\left(\dfrac{6}{17}-\dfrac{47}{53}\right)\)
\(=\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{47}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}-\dfrac{17}{3}.\dfrac{47}{53}\)
\(=\left(\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{17}{3}.\dfrac{47}{53}\right)+\left(\dfrac{-47}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}\right)\)
\(=\left(1-1\right)+\left(-1+2\right)\)
\(=0+1\)
\(=1\)
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c. TA CÓ:
\(\frac{33}{132}=\frac{1}{4}\) mà \(\frac{33}{131}>\frac{33}{132}\) suy ra \(\frac{33}{131}>\frac{1}{4}\) (1)
\(\frac{53}{212}=\frac{1}{4}\) mà \(\frac{53}{217}
d. TA CÓ:
\(\frac{41}{91}=\frac{410}{910}=1-\frac{500}{910}\); \(\frac{411}{911}=1-\frac{500}{911}\)
TA THẤY VÌ \(\frac{500}{910}>\frac{500}{911}\) NÊN \(1-\frac{500}{910}
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a)(1/4+3/4)-(5/13+8/13)+2/11=1-1+2/11=2/11
b)(21/31+10/31)-16/7+(44/53+9/53)=1+1-16/7=14/7-16/7=-2/7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1)\) \(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
\(2)\) Đặt \(A=4+2^2+2^4+...+2^{20}\)
\(4A=2^4+2^4+2^6+...+2^{22}\)
\(4A-A=\left(2^4+2^4+2^6+...+2^{22}\right)+\left(2^2+2^2+2^4+...+2^{20}\right)\)
\(3A=2^4+2^{22}-2^2-2^2\)
\(3A=2^{22}+2^4-2^3\)
\(A=\frac{2^{22}+2^4-2^3}{3}\)
\(3)\) \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\) ( bạn ghi đầy đủ ra nhé ở đây mk viết "..." cho nhanh )
\(=\)\(5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=\)\(5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=\)\(5\left(1-\frac{1}{31}\right)\)
\(=\)\(5.\frac{30}{31}\)
\(=\)\(\frac{150}{31}\)
Chúc bạn học tốt ~
Ta có:
\(\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
\(\left(800+53\right).53+\left(900-47\right).47\)
\(=853.53+853.47\)
\(=853.\left(53+47\right)\)
\(=853.100\)
\(=85300\)
(800+53).53+(900-47).47=853.53+853.47=853.(53+47)=853.100=85300