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a ) Ta có
\(\frac{29}{33}>\frac{29}{37}\)( đồng tử khác mẫu )
\(\frac{22}{37}< \frac{29}{37}\)( đồng mẫu khác tử )
=> \(\frac{29}{33}>\frac{29}{37}>\frac{22}{37}\)
b ) \(\frac{163}{257}< \frac{163}{221}\)
\(\frac{162}{257}>\frac{149}{257}\)
\(\Rightarrow\frac{163}{221}>\frac{163}{257}>\frac{149}{257}\)
a) ta có: \(\frac{22}{37}< \frac{29}{37}\)
\(\frac{29}{33}>\frac{29}{37}\)
\(\Rightarrow\frac{22}{37}< \frac{29}{37}< \frac{29}{33}\)
b) ta có: \(\frac{163}{257}>\frac{149}{257}\)
\(\frac{163}{221}>\frac{163}{257}\)
\(\Rightarrow\frac{163}{221}>\frac{163}{257}>\frac{149}{257}\)
a) (100 - 97)2 + 720 : 718 + 20130
= 32 + 72 + 1
= 9 + 49 + 1
= 59
b) 37 . 29 +29 . 62 + 29
= 29 . ( 37 + 62 ) + 29
= 29 . 99 + 29
= 2871 + 29 = 2900
c) ( -2013 ) + 1908 + 2013 + ( -908 )
= ( -105 ) + 2013 - 908
= 1908 - 908 = 1000
d) (-35 ) - ( 12 - 17 )
= ( -35 ) - ( -5 )
= - (35 - 5)
= -30
a ) 31 x ( - 125 ) + 225 x 31
= 31 x [ ( - 125 ) + 225 ]
= 31 x 100
= 3100
b ) 26 x ( - 125 ) - 125 x ( - 36 )
= ( - 26 ) x 125 - 125 x ( - 36 )
= 125 x [ ( - 26 ) - ( - 36 ) ]
= 125 x 10
= 1250
a) 31 × (-125) + 225 × 31
= 31 x ( -125 + 225 )
= 31 x 100
= 3100
b) 26 × (-125) – 125 × (-36)
= -125 ( 26 - 36 )
= 125x 10
= 1250
d) 17 × (-37) – 23 × 37 – 46 × (-37)
= -629 - 851 + 1702
= 1480
b) (-17) . 37 - 23 . 37 - (-46) . 37
= 37 . {(-17) - 23 - (-46)}
= 37 . 6 = 222
a) 29.(-13) +27.(-29)+(-14).-(29)
=(-29).13+ 27.(-29)+(-14).(-29)
=(-29).[13+27+(-14)]
=(-29).26
=-754
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 :<<
Bài 1:
a) Ta có: \(13A=\dfrac{13^{16}+13}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)
\(13B=\dfrac{13^{17}+13}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)
Vì \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(1999C=\dfrac{1999^{2000}+1999}{1999^{2000}+1}=1+\dfrac{1998}{1999^{2000}+1}\)
\(1999D=\dfrac{1999^{1999}+1999}{1999^{1999}+1}=1+\dfrac{1998}{1999^{1999}+1}\)
\(\dfrac{1998}{1999^{2000}+1}< \dfrac{1998}{1999^{1999}+1}\Rightarrow1+\dfrac{1998}{1999^{2000}+1}< 1+\dfrac{1999}{1999^{1999}+1}\)
\(\Rightarrow1999C< 1999D\)
\(\Rightarrow C< D\)
Vậy C < D
77
viết hộ mình lời giải nhé