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 minh rằng: 1/20*23+1/23*26+1/26*29+....+1/77*80 <1/79
\(\frac{1}{20\cdot23}+\frac{1}{23\cdot26}+\frac{1}{26\cdot29}+...+\frac{1}{77\cdot80}\)
\(< \frac{1}{3}\left[\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+\frac{3}{26\cdot29}+...+\frac{3}{77\cdot80}\right]\)
\(< \frac{1}{3}\left[\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\left[\frac{1}{20}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\left[\frac{4}{80}-\frac{1}{80}\right]\)
\(< \frac{1}{3}\cdot\frac{3}{80}=\frac{1}{80}< \frac{1}{79}(đpcm)\)
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:
S2= 32 phần 20*23 + 32 phần 23*26 + 32 phần 2*29 + . . . + 32 phần 77*80 < 1 phần 8
[ 22^32(32^23-23^22) - 32^23(22^32-23^22) ] : 23^22(32^23-22^32)
B =3mu 2/20*23 +3mu 2/23*26 +....+3mu2/ 77*80
Tính nhanh:
C=23^22.(32^23-22^32)/22^32.(32^23-23^22)-32^23.(22^32-23^22)
tính nhanh [22^32.(32^23-23^22)-32^28.(22^32-23^22)]:23^22.(32^23-22^32)
Nguyễn Quang Thành trả lời cụ thể đi
Chứng minh rằng:
32/20×23+32/23×36+.....+32/77×80<1
Xin hãy giúp tôi
Ta có : \(\frac{3^2}{20\cdot23}+\frac{3^2}{23\cdot26}+...+\frac{3^2}{77\cdot80}=\frac{1}{3}\cdot\left(\frac{1}{20}-\frac{1}{80}\right)=\frac{1}{3}\cdot\frac{3}{80}=\frac{1}{80}< 1\) ( đpcm )