B =3mu 2/20*23 +3mu 2/23*26 +....+3mu2/ 77*80
cho A=2+2mu2 + 2mu3+..+2mu20tim chhu so tan cung chua A
B=1+3+3mu2 +3mu 3+...+3mu 30chung minh B khong phai la so chinh phuong
chu y so chinh phong la binh phuong
tim a,b thuoc N10a + 168=b2
Tổng A có 20 số, nhóm 4 số vào 1 nhóm thì vừa hết.
Ta có;
A = (2 + 22 + 23 + 24) + (25 + 26 + 27 + 28) +......+ (217 + 218 + 219 + 220)
= (2 + 22 + 23 + 24) + 24(2 + 22 + 23 + 24) + ...... + 216(2 + 22 + 23 + 24)
= 30 + 24.30 + ......+ 216.30
= 30(1 + 24 + .......+ 216) = ....0
=> A có chữ số tận cùng là 0.
B=3+3mu2+3mu3+.....3mu 2000
la boi cua so tu nhien nao
A=3+3mu2+3mu3 +....3mu9+3mu 10. chứng minh A chia het cho 4
chứng minh rằng: 32/20*23+32/23*26+...+32/77*80<1
(*: là dấu nhân )
Chứng minh rằng 1/20*23+1/23*26+1/26*29+...+1/77*80<1/79
Đặt : \(A=\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77.80}\)
\(\Rightarrow3A=\frac{3}{20.23}+\frac{3}{23.26}+\frac{3}{26.29}+...+\frac{3}{77.80}\)
\(\Rightarrow3A=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\)
\(\Rightarrow3A=\frac{1}{20}+\left(\frac{1}{23}-\frac{1}{23}\right)+\left(\frac{1}{26}-\frac{1}{26}\right)+...+\left(\frac{1}{77}-\frac{1}{77}\right)-\frac{1}{80}\)
\(\Rightarrow3A=\frac{1}{20}-\frac{1}{80}\)
\(\Rightarrow3A=\frac{3}{80}\)
\(\Rightarrow A=\frac{3}{80}:3\)
\(\Rightarrow A=\frac{1}{80}\)
Vì 80 > 79 nên \(\frac{1}{80}< \frac{1}{79}\)hay \(A< \frac{1}{79}\)
~ Hok tốt ~
chứng tỏ rằng 1/20*23+1/23*26+1/26*29.........+1/77*80<1/9
Ta có :
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\left(\frac{3}{80}< 1\right)\)
\(\Leftrightarrow\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}< \frac{1}{3}\left(đpcm\right)\)
\(M=\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77x80}\)
\(M=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\)
\(M=\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)
\(\frac{3}{80}=\frac{3x9}{80x9}=\frac{27}{720};\frac{1}{9}=\frac{1x80}{9x80}=\frac{80}{720}\)
Vì \(\frac{27}{720}< \frac{80}{720}\Rightarrow\frac{3}{80}< \frac{1}{9}\Rightarrow M< \frac{1}{9}\)
#~Will~be~Pens~#
Hoàng Nguyên Hiếu:Sai rồi nha bạn
\(\frac{1}{20.23}=\frac{1}{20}-\frac{1}{23}\Leftrightarrow23-20=1\)
-.-
Chứng minh rằng: 1/20*23+1/23*26+1/26*29+....+1/77*80 <1/79
\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+\frac{1}{26\cdot29}+...+\frac{1}{77\cdot80}\)
\(< \frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+\frac{3}{26\cdot29}+...+\frac{3}{77\cdot80}\right]\)
\(< \frac{1}{3}\left[\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\left[\frac{1}{20}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\cdot\frac{3}{80}=\frac{1}{80}< \frac{1}{79}(đpcm)\)
2mu2.5mu2-(2008 mu0+8 ):3mu 2
2mu3.5mu2 -(2014 mu0+8):3mu2+ 0 mu23
2.3mu3+143:(2016mu0.6mu2 -625:25)
minh can ngay bay gio giup minh voi
`2^2 *5^2 -(2008^0 +8):3^2`
`=4*25-(1+8):9`
`=100-9:9`
`=100-1`
`=99`
`2^3 *5^2 -(2014^0 +8):3^2 +0^23`
`=8*25-(1+8):9+0`
`=200-9:9`
`=200-1`
`199`
`2*3^3 +143:(2016^0 * 6^2 -625:25)`
`=2*27+143:(1*36-25)`
`=54+143:11`
`=54+13`
`=67`
HELP !
B=5/20*23+5/23*26+...+5/77*78 cmr B<1/2
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