chứng minh rằng căn 50+căn2026>50
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27 tháng 2 2016
\(\sqrt{99}<\sqrt{100}=10\)
\(\sqrt{50}+\sqrt{10}\)\(>\sqrt{49}+\sqrt{9}=7+3=10\)
Vậy \(\sqrt{50}+\sqrt{10}>\sqrt{99}\)
2 tháng 4 2016
CM:căn 50 + căn 49 >14
căn 50+7>14
căn 50>7
mà căn 50> căn 49=7
=> căn 50 + căn 49>14
k cho mình nha
H
3
28 tháng 6 2021
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)(đpcm)
SI
28 tháng 6 2021
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\) (đpcm)
4 tháng 7 2016
1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)
=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)
=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25
=1/26+1/27+...+1/50 (đpcm)
24 tháng 8 2023
Ta có:
50! - 40! = 40!(41.42.43.44.45.46.47.48.49.50 - 1)
Mà 40! = 40.39.38.37...2.1 nên 40! chia hết cho 37
Suy ra 50! - 40! chia hết cho 37
TH
3
N
17 tháng 10 2018
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{49}+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{40}+\frac{1}{50}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
Vậy .....(tự kết luận)
T
17 tháng 10 2018
CMR: \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
\(VT=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}^{\left(đpcm\right)}\)
LT
1
9 tháng 10 2016
Ta biến đổi vế phải :
1-1/2+1/3-1/4+.....+1/49-1/50
=(1+1/3+1/5+....+1/49)-(1/2+1/4+1/6+.......+1/50)
=(1+1/2+1/3+.....+1/49+1/50)-2(1/2+1/4+1/6+......+1/50)
=(1+1/2+...+1/50)-(1+1/2+1/3+....+1/25)
=1/26+1/27+.......+1/50
Vậy 1/26+1/27+1/28+.....+1/50=1-1/2+1/3-1/4+......+1/49-1/50
Mình không bấm phân số được mong mấy bạn thông cảm
PN
9
28 tháng 7 2015
1/26+1/27+1/28+...+1/49+1/50=1-1/2+1/3-1...
<=>2/26+2/28+2/30+...+2/50=1-1/2+1/3-1...
<=>1/13+1/14+1/15+...+1/25=1-1/2+1/3-1...
<=>2/14+2/16+2/18+...2/24=1-1/2+1/3-1/...
<=>1/7+1/8+1/9+...+1/12=1-1/2+1/3-1/4+...
<=>2/8+2/10+2/12=1-1/2+1/3-1/4+1/5-1/6
<=>1/4+1/5+1/6=1-1/2+1/3-1/4+1/5-1/6
<=>2/4+2/6=1-1/2+1/3
<=>1/2+1/3=1-1/2+1/3
<=>2/2=1
28 tháng 7 2015
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
=>đpcm