tính hợp lí a 258 - 173 - 27 + 42 - 50
b - 1962 + 1963 - 1946 - 1146+101
c 1-2+3-4+...+199- 200
Những câu hỏi liên quan
-1962+1963+1246+1146+101
1379-(27+379)
4971-(917-289)
-1962+1963+1246+1146+101
= 1 + 1246 + 1146 + 101
= 1247 + 1146 + 101
= 2393 + 101
= 2494
1379-(27+379)
= 1379 - 406
= 1785
4971-(917-289)
= 4971 - 628
= 4343
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-1962+1963+1246+1146+101=2496
1379-(27+379)=973
4971-(917-289)=4343
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Tính tông sau một cách hợp lí
A= 1-3+6-7+.....+97-99+101
B= -1-2-3-4-5 -........-199-200-201-202
435-43-483-57+383-415
1316-85+317-1216-315
-1962+1963-1246+1146+101
Mọi người ơi giải hộ mình mấy bài này với
Tính tỉ số A/B biết:
A= 1/1*2+1/3*4+1/5*6+...+1/199+200
B= 1/101*200+1/102*199+...+1/200*101
Tính tỉ số A/B biết:
A= 1/1*2+1/3*4+1/5*6+...+1/199+200
B= 1/101*200+1/102*199+...+1/200*101
A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Lại có B = \(\frac{1}{101.200}+\frac{1}{102.199}+...+\frac{1}{200.101}\)
=> 301B = \(\frac{301}{101.200}+\frac{301}{102.199}+...+\frac{301}{200.101}\)
=> 301B = \(\frac{1}{101}+\frac{1}{200}+\frac{1}{102}+\frac{1}{199}+...+\frac{1}{200}+\frac{1}{101}=2\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
=> B = \(\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}{\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}=\frac{1}{\frac{2}{301}}=\frac{301}{2}=150,5\)
Bài 1. Chứng minh rằng:
A = 2/3 . 4/5 . ... . 4998/4999 < 0,02
Bài 2. Chứng minh rằng:
a) 1/26 + 1/27 + ... + 1/56 = 99/50 - 97/49 + ... + 7/4 - 5/3 + 3/2 - 1
b) 1- 1/2 + 1/3 - 1/4 + ... + 1/199 - 1/200 = 1/101 + 1/102 + ... + 1/200
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)( đpcm )
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Tính các tổng sau:
a) A = 1*2+2*3+3*4+...+2014*2015
b) B = 101^2+102^2+...+199^2+200^2
c) C = 1*3+2*4+3*5+4*6+...+99*101+100*102
cho mi sửa lại:
\(a) A = 1^2+2^3+3^4+...+2014^{2015} b) B = 101^2+102^2+...+199^2+200^2 c) C = 1^3+2^4+3^5+4^6+...+99^{101}+100^{102}\)
dấu 8 là nhân còn dấu ^ là mũ ạ
cho a=1/1*2+1/3*4+1/5*6+...+1/199*200
b=1/101+1/102+...+1/200
tính a/b
Ta có: \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
\(\Rightarrow A=B\)
Khi đó, \(\frac{A}{B}=1\)
Tính tổng sau bằng hợp lí :-1-2-3-4-...-199-200-201-202
Đặt A=1+2+3+...+201+202
A có: (202-1)+1=202(số hạng)
A=(202+1)*202/2=20503
=>-A=-(1+2+3+...+201+202)=-1-2-3-4-...-199-200-201-202=-20503
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