Cmr: 1/3! + 1/4! + 1/5! +... +1/2000! < 1/2
Những câu hỏi liên quan
1. CMR:1/1.2+1/3.4+1/5.6+....+1/49.50=1/26+1/27+.......+1/50
2.A=1/1.2+1/2.3+.....+1/99.100.CMR:7/12<A<5/6
3.tim x
a.x+1/10+ x+1/12 +x+1/14 + x+1/16 + x+1/18 + x+1/20
b.x+1/2000 + x+2/1999 =x+3/1998 + x+4/1977
giup minh nha cac ban
1/1.2+1/3.4+1/5.6+...+1/49.50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=1/1+1/2+1/3+1/4+...+1/49+1/50-2(1/2+1/4+1/6+...+1/50)
=1/1+1/2+1/3+1/4+...+1/49+1/50-(1/1+1/2+1/3+1/4+...+1/25)
=1/26+1/27+...+1/50=1/26+1/27+...+1/50(đpcm)
b. 1/1-1/2+1/3-1/4+...+1/99-1/100=99/100
7/12=175/300; 5/6=10/12=250/300; 99/100=297/300
(hình như khúc này đề bài sai hả bạn) bạn tự tính ra nhé
bài 2: a.x+1/10+x/12+x/14+...x+1/20
(x+x+x...+x)+(1/10+1/12+...+1/20)
ko có kết quả sao tìm x được bạn:[
b.x+1/2000+x+2/1999=x+3/1998+x+4/1997
x+1/2000+x+2/1999=x+3/1998+x+4/1997
(x+1/2000+1)+(x+2/1999+1)=(x+3/1998+1)+(x+4/1997+1)
x+2002/2000+x+2002/1999=x+2002/1998+x+2002/1997
x+2002(1/2000+1/1999)=(x+2002)(1/1998+1/1997)
=>(1/2000+1/1999)=(1/1998+1/1997)
x+2002(1/2000+1/1999)-(x+2002)(1/1998+1/1997)=0
(x+2002)(1/2000+1/1999-1/1998-1/1997)=0
(x+2002).0=0
(x+2002)=0
x =0-2002=-2002
Chúc bạn học tốt.
Đúng 0
Bình luận (1)
Bài 1: CMR:a, A1/2+1/3^2+1/3^3+...+1/3^991/2b, B1/2+(1/2)^2+(1/2)^4+...+(1/2)^98+(1/2)^991Bài 2 CMR:a, 7^6+7^5-7^4 chia hết cho 55b,3^n+2-2^n+2+3^n-2^n chi hết cho 10 ( với mọi số nguyên dương a)c,43^43-17^17 chia hết cho 10d, 23^401+38^201-2^433 chia hết cho 5Bài 3 Tìm x, biết: / / là dấu GTTĐ a, /x-5/x+3b, /x-5//x+3/c, /x-1/+/x-3/2xd, (2x-3)^24/9 e, (2x-1)^3-8f, 35-x/x+143/5g, 0,4:xx:0,9h, x+4/2000+x3/2001x-3/2000+x-4/1999i, x-1/2002+x-2/2001x-3/2000+x-4/1999k, (2x+1/3)(3x-1/4)(4x+1/5)...
Đọc tiếp
Bài 1: CMR:
a, A=1/2+1/3^2+1/3^3+...+1/3^99<1/2
b, B=1/2+(1/2)^2+(1/2)^4+...+(1/2)^98+(1/2)^99<1
Bài 2 CMR:
a, 7^6+7^5-7^4 chia hết cho 55
b,3^n+2-2^n+2+3^n-2^n chi hết cho 10 ( với mọi số nguyên dương a)
c,43^43-17^17 chia hết cho 10
d, 23^401+38^201-2^433 chia hết cho 5
Bài 3 Tìm x, biết: / / là dấu GTTĐ
a, /x-5/=x+3
b, /x-5/=/x+3/
c, /x-1/+/x-3/=2x
d, (2x-3)^2=4/9
e, (2x-1)^3=-8
f, 35-x/x+14=3/5
g, 0,4:x=x:0,9
h, x+4/2000+x=3/2001=x-3/2000+x-4/1999
i, x-1/2002+x-2/2001=x-3/2000+x-4/1999
k, (2x+1/3)(3x-1/4)(4x+1/5)
n, 5^x+5^(x+2)= 650
m, 3^(x-1) +5.3^(x-1)=162
Bài 1: CMR 3/1^2*2^2 + 5/2^2*3^2 + 7/3^2*4^2 + ....... + 19/9^2*10^2 bé hơn 1
Bài 2: CMR 1/3 + 2/3^2 Bài 1: CMR 3/1^2*2^2 + 5/2^2*3^2 + 7/3^2*4^2 + ....... + 19/9^2*10^2 bé hơn 3/4
Bài 3: Cho A= 1/1*2 + 1/3*4 + 1/5*6 + .... + 1/99*100. CMR 7/12 < A < 5/6
ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều
CMR \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}=\dfrac{1}{1002}+...+\dfrac{1}{2002}\)
đặt \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2001}-\dfrac{1}{2002}\\ Q=\dfrac{1}{1002}+...+\dfrac{1}{2002}\)
ta có:
\(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2001}-\dfrac{1}{2002}\\ \Rightarrow P=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2001}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)\)\(\Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)\\ \Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2002}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1001}\right)\\ \Rightarrow P=\dfrac{1}{1002}+...+\dfrac{1}{2002}\\ \Rightarrow P=Q\)\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2001}-\dfrac{1}{2002}=\dfrac{1}{1002}+...+\dfrac{1}{2002}\left(đpcm\right)\)
Đúng 0
Bình luận (1)
C=1/4^2+1/6^2+1/8^2+......+1/2000^2 cmr C<665/402
1)2/5+x:5/7=1/3
CMR: 2)B=1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2<1
3)CMR: S=3^2+3^3+...+3^101 chia hết cho 120
4)Cho S=5+5^2+5^3+...+5^2006
a) tính S
b)CMR S chia hết cho 6, và S chia hết cho 30
5) tìm số tự nhiên n sao cho 4n-5 chia hết cho 2n-1
Bài 1 : Tìm x :
1) 36^2-49=0
2) x^3-16x=0
3) (x-1)*(x+2)-x-2=0
4) 3x^3-27x=0
5) x^2*(x+1)+2x*(x+1)=0
6) x*(2x-3)-2*(3-2x)=0
Bài 2 : Toán chia hết :
a) CMR 8^5+2^11chia hết cho 17
b) CMR 69^2-69.5chia hết cho 32
c) CMR 328^3+172^3 chia hết cho 2000
d) CMR 19^19+69^19 chia hết cho 44
e) CMR hiệu các bình phương của hai số lẻ liên tiếp chia hết cho 8
1/M=1/1+2+3+1/1+2+3+4+1/1+2+3+4+5+...+1/1+2+3+4+..+59
cmr M>2/3
CMR(1/1*2+1/2*3+1/3*4+1/4*5+...+1/99*100):(1/51+1/52+1/53+...+1/100) = 1
Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
=1
Đúng 0
Bình luận (0)