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Ta có : \(\frac{2003.2004-1}{2003.2004}=\frac{2003.2004}{2003.2004}-\frac{1}{2003.2004}=1-\frac{1}{2003.2004}\)
\(\frac{2004.2005-1}{2004.2005}=\frac{2004.2005}{2004.2005}-\frac{1}{2004.2005}=1-\frac{1}{2004.2005}\)
Vì \(\frac{1}{2003.2004}>\frac{1}{2004.2005}\)
Nên : \(\frac{2003.2004-1}{2003.2004}< \frac{2004.2005-1}{2004.2005}\)
Bài 1:
5; (-23) + 105
= 105 - 23
= 82
6; 78 + (-123)
= 78 - 123
= - (123 - 78)
= - 45
bài1
1)2763 + 152 = 2915
2)-7 +(-14)
=-(14 +7)
=-21
a) Ta có:
+) \(\dfrac{1}{2}=\dfrac{3}{6}\)
+) \(\dfrac{1}{3}=\dfrac{2}{6}\)
+) \(\dfrac{2}{3}=\dfrac{4}{6}\)
=> \(\dfrac{2}{6}< \dfrac{3}{6}< \dfrac{4}{6}\)
hay \(\dfrac{1}{3}< \dfrac{1}{2}< \dfrac{2}{3}\)
b) Ta có:
+) \(\dfrac{4}{9}=\dfrac{56}{126}\)
+) \(-\dfrac{1}{2}=-\dfrac{63}{126}\)
+) \(\dfrac{3}{7}=\dfrac{54}{126}\)
=> \(-\dfrac{63}{126}< \dfrac{54}{126}< \dfrac{56}{126}\)
hay \(-\dfrac{1}{2}< \dfrac{3}{7}< \dfrac{4}{9}\)
c) Ta có:
+) \(\dfrac{27}{82}=\dfrac{2025}{6150}\)
+) \(\dfrac{26}{75}=\dfrac{2132}{6150}\)
=> \(\dfrac{2025}{6150}< \dfrac{2132}{6150}\)
hay \(\dfrac{27}{82}< \dfrac{26}{75}\)
d) Ta có:
+) \(-\dfrac{49}{78}=-\dfrac{4655}{7410}\)
+) \(-\dfrac{64}{95}=-\dfrac{4992}{7410}\)
=> \(-\dfrac{4665}{7410}>-\dfrac{4992}{7410}\)
hay \(-\dfrac{49}{78}>-\dfrac{64}{95}\)
Bài 1 : bạn tự làm
Bài 2 :
1) (15+37)+(52-37-17) = 15 + 37 +52 - 37 - 17 = 50
2) (38 - 42 +14 ) - (25 - 27 -15 ) = 38 -42 +14 -25 +27 +15 = 27
3) - (21 - 32) - (-12 +32) = -21 +32 +12 -32 = -9
4) - (12 +21 -23) - (23 -21 +10) = -12 -21 +23 -23 +21 -10 = -22
5) (57-725) - (605 -53 ) = 57 -725 - 605 +53 = -1220
6) (55 +45 +15) - (15 -55 +45) = 55 +45 +15 -15 +55 -45 = 110
7) (35 +75) + (345 -35 -75) = 35 +75 +345 -35 -75 = 345
8) (2002 -79 +15) - (-79 +15 )= 2002 -79 +15 +79 -15 = 2002
9) -(515 -80 +91) - (2003+80-91) =-515+80-91-2003-80+91=-2518
10) 25-(-17) +24 -12 = 25 +17 +24 -12 = 54
11) 235 - (34 +135) -100 = 235 - 34 -135 -100 = -34
12) (13 +49 ) - (13 -135 +49) = 13 +49 -13 +135 -49 =135
13) (18 +29) + (158 -18 -29) = 18 +29 +158 -18 -29 = 158
Bài 3:
1) (-37)+14+26+37 = (14+26)+37-37 = 40
2) (-24) +10+6+24 = (10 +6) +24 -24 = 16
3) 15 +23 + (-25 )+(-23) = (15-25)+(23-23) = -10
4) 60+33+(-50) + (-33) = (60-50)+(33-33) = 10
5) (-16)+(-209)+(-14) +209 = (-16-14)+(209-209) = -30
6) (-12)+(-13)+36+(-11) = (-12-13-11)+36 = (-36)+36 = 0
7) -16 +24 +16 -34 = (-16 +16)+(24-34) = -10
8) 25 +37-48-25-37 = (25-25)+(37-37)-48 = -48
9) 2575+37-2576-29 = (2575-2576)+(37-29)= -1+8 = 7
10) 34+35+36+37-14-15-16-17= (34-14)+(35-15)+(36-16)+(37-17)
= 20+20+20+20 = 80
11) 4573+46-4573+35-16-5= (4573-4573)+(46-16)+(35-5)
= 30 +30 = 60
12) 32+34+36+38-10-12-14-16-18
= (32-12)+(34-14)+(36-16)+(38-18)-10
= 20+20+20+20-10 = 70
a/
\(37^{1320}=\left(37^2\right)^{660}=1369^{660}\)
\(11^{1979}< 11^{1980}=\left(11^3\right)^{660}=1331^{660}\)
\(\Rightarrow1363^{660}>1331^{660}\Rightarrow37^{1320}>11^{1979}\)
b/
\(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
\(\Rightarrow27^{11}>81^8\)
d/
\(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 9^{21}< 11^{21}\)
e/ \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
\(\Rightarrow5^{36}>11^{24}\)
g/ \(21^{15}=3^{15}.7^{15}\)
\(27.49^8=3^3.\left(7^2\right)^8=3^3.7^{16}\)
\(\frac{21^{15}}{27.49^8}=\frac{3^{15}.7^{15}}{3^3.7^{16}}=\frac{3^{12}}{7}>1\Rightarrow21^{15}>27.49^8\)
f/ \(199^{20}=\left(199^4\right)^5\)
\(2003^{15}=\left(2003^3\right)^5\)
\(2003^5>1990^5\)
\(\frac{1990^5}{199^4}=\frac{199^5.10^5}{199^4}=199.10^5>1\)
\(\Rightarrow2003^5>1990^5>199^4\Rightarrow2003^{15}>199^{20}\)
Bài 1 :
a) \(\dfrac{42}{43}=1-\dfrac{1}{43}\)
\(\dfrac{58}{59}=1-\dfrac{1}{59}\)
Mà \(\dfrac{1}{43}>\dfrac{1}{59}\Leftrightarrow\dfrac{42}{43}< \dfrac{58}{59}\)
b) \(\dfrac{18}{31}>\dfrac{15}{31}>\dfrac{15}{37}\)
\(\Leftrightarrow\dfrac{18}{31}>\dfrac{15}{37}\)
c) \(\dfrac{53}{57}=1-\dfrac{4}{57}\)
\(\dfrac{531}{517}=1-\dfrac{40}{517}\)
Mà \(\dfrac{4}{57}=\dfrac{40}{570}>\dfrac{40}{517}\)
\(\Leftrightarrow\dfrac{53}{57}< \dfrac{531}{517}\)