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Trên máy mk hiển thị , câu hỏi này 4 phút nữa mới chính thức xuất hiện ,,, máy bị j hay do câu hỏi ak ??
a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)
\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)
Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)
\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)
\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)
\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)
Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)
\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)
\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
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
S=\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}<\frac{4}{10}+\frac{4}{10}+\frac{4}{10}+\frac{4}{10}+\frac{4}{10}\)
=\(\frac{4}{10}\cdot5=2=>S<2\)
S=\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}<\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}\)
=\(\frac{3}{15}\cdot5=1=>S>1\)
Vậy 1<S<2
nhớ k với nhé
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
\(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{70}\)
\(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}\right)\)
\(+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)\)
\(\Rightarrow A< \frac{1}{10}\cdot10+\frac{1}{20}\cdot10+\frac{1}{30}\cdot10+...+\frac{1}{60}\cdot10\)
\(A< 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{6}\)
\(A< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\left(\frac{1}{4}+\frac{1}{5}\right)\)
\(A< 2+0,45< 2,5\)
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(A>\left(\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}\right)+\left(\frac{1}{30}+...+\frac{1}{30}\right)+...+\left(\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\right)\)
\(A>\frac{1}{2}+\frac{1}{3}+..+\frac{1}{7}\)
\(A>\frac{223}{140}>\frac{4}{3}\)