1+20+(-20)*1+23= ???
K
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LQ
1
16 tháng 3 2018
\(\frac{4}{3}\cdot7+\frac{5}{7}\cdot12+\frac{1}{12}\cdot13+\frac{7}{13}\cdot20+\frac{3}{20}\cdot23+\frac{9}{23}\cdot32\)
BẠN VIÊT THẾ NÀY THÌ MN DỄ HIỂU HƠN NHÉ
VH
3
12 tháng 7 2023
a) 23x51+75x23-23x25=23x(51+75-26)=23x100=2300
b) 1+2+3+...+20=\(\dfrac{\left(20+1\right)\text{x}20}{2}\)=210
H9
HT.Phong (9A5)
CTVHS
12 tháng 7 2023
a) \(23\times51+75\times23-23\times26\)
\(=23\times\left(51+75-26\right)\)
\(=23\times\left(126-26\right)\)
\(=23\times100\)
\(=2300\)
b) \(1+2+...+20\)
\(=\left(20+1\right)+\left(19+2\right)+\left(18+3\right)+\left(17+4\right)+...+\left(11+10\right)\)
\(=21+21+21+...+21\) (10 số 21)
\(=2100\)
S
1
13 tháng 5 2019
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 ~
KT
4
KN
4 tháng 5 2019
Ta có :
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\left(\frac{3}{80}< 1\right)\)
\(\Leftrightarrow\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}< \frac{1}{3}\left(đpcm\right)\)
T
4 tháng 5 2019
\(M=\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77x80}\)
\(M=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\)
\(M=\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)
\(\frac{3}{80}=\frac{3x9}{80x9}=\frac{27}{720};\frac{1}{9}=\frac{1x80}{9x80}=\frac{80}{720}\)
Vì \(\frac{27}{720}< \frac{80}{720}\Rightarrow\frac{3}{80}< \frac{1}{9}\Rightarrow M< \frac{1}{9}\)
#~Will~be~Pens~#
NH
1
13 tháng 5 2019
\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+\frac{1}{26\cdot29}+...+\frac{1}{77\cdot80}\)
\(< \frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+\frac{3}{26\cdot29}+...+\frac{3}{77\cdot80}\right]\)
\(< \frac{1}{3}\left[\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\left[\frac{1}{20}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\cdot\frac{3}{80}=\frac{1}{80}< \frac{1}{79}(đpcm)\)
NH
1
28 tháng 7 2017
Gọi \(A=\dfrac{3^{17}+1}{3^{20}+1}\) và \(B=\dfrac{3^{20}+1}{3^{23}+1}\)
Ta Có: \(A=\dfrac{3^{17}+1}{3^{20}+1}=\left(\dfrac{3^{17}+1}{3^{20}+1}\right).\dfrac{3^3}{3^3}=\dfrac{3^{20}+27}{3^{23}+27}\)
Ta lại có: \(1-A=1-\dfrac{3^{20}+27}{3^{23}+27}=\dfrac{3^{23}+27}{3^{23}+27}-\dfrac{3^{20}+27}{3^{23}+27}=\dfrac{3^{23}-3^{20}}{3^{23}+27}\)
\(1-B=1-\dfrac{3^{20}+1}{3^{23}+1}=\dfrac{3^{23}+1}{3^{23}+1}-\dfrac{3^{20}+1}{3^{23}+1}=\dfrac{3^{23}+3^{20}}{3^{23}+1}\)
Vì: \(\dfrac{3^{23}-3^{20}}{3^{23}+27}< \dfrac{3^{23}-3^{20}}{3^{23}+1}\Rightarrow A>B\)
Vậy \(\dfrac{3^{17}+1}{3^{20}+1}>\dfrac{3^{20}+1}{3^{23}+1}\)
1 + 20 + (-20) x 1 + 23
= 1 + 20 - 20 + 23
= 1 + 0 + 23
= 1 + 23
= 24
=(1+23)+[20+(-20)]
=24+0
=24