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\(\dfrac{3^2}{20.23}\)+\(\dfrac{3^2}{23.26}\)+...+\(\dfrac{3^2}{77.80}\)
=> \(\dfrac{9}{20.23}+...+\dfrac{9}{77.80}\)
= 9.\(\left(\dfrac{1}{20.23}+...+\dfrac{1}{77.80}\right)\)
\(=9.\left(\dfrac{1}{20.3}-\dfrac{1}{23.3}+\dfrac{1}{23.3}-\dfrac{1}{26.3}+...+\dfrac{1}{77.3}-\dfrac{1}{80.3}\right)\)= \(9.\left(\dfrac{1}{20.3}-\dfrac{1}{80.3}\right)\)
\(=9.\dfrac{1}{80}\)=\(\dfrac{9}{80}=0,1125< 1.\)
3^2= 9
Vậy thì sẽ là:
9/ 20.23+ 9/ 23.26+...9/77.80
cách nhau 3 bỏ 3 ra ngoài
= 3(3/20.23+...3/77.80)
=3(3/20-3/23+3/23-3/26+.....+3/77-3/80)
=3(3/20-3/80)
=3. 9/80
=27/80<1
32=9
\(\frac{3^2}{20.23}\)+\(\frac{3^2}{23.26}\)+...+\(\frac{3^2}{77.80}\)
=\(\frac{9}{20.23}\)+\(\frac{9}{23.26}\)+...+\(\frac{9}{77.80}\)
=3(\(\frac{3}{20.23}\)+\(\frac{3}{23.26}\)+...+\(\frac{3}{77.80}\))
=3(\(\frac{1}{20}\)-\(\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\))
=3(\(\frac{1}{20}-\frac{1}{80}\))
=3(\(\frac{4}{80}-\frac{1}{80}\))
=3.\(\frac{3}{80}\)
=\(\frac{9}{80}\)<1
Vậy\(\frac{9}{80}< 1\)
tính chứ ko phải chứng minh đâu bạn?
\(=3^2\left(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\right)\)
\(=3^2.\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{9}{80}\)
Đặt \(A=\frac{3^2}{20\cdot23}+\frac{3^2}{23\cdot26}+\frac{....3^2}{77\cdot80}\)
\(A=3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+....+\frac{3}{77\cdot80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(A=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(A=3\cdot\frac{3}{80}\)
\(A=\frac{9}{80}\)
Ta có:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\Rightarrowđpcm\)
Vậy \(A< \frac{3}{4}\)
Cho A = 1/32 + 1/33 + 1/34 + ... + 1/39
=>3A=1/3+1/32+1/33+...+1/38
=>3A-A=1/3+1/32+1/33+...+1/38-1/32-1/33-1/34-...-1/39
=>2A=1/3-1/39
=>\(A=\frac{\frac{1}{3}-\frac{1}{3^9}}{2}\)<1
Vậy A<1
\(\frac{A}{3}=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)
\(\frac{A}{3}=\frac{23-20}{20.23}+\frac{26-23}{23.26}+...+\frac{80-77}{77.80}\)
\(\frac{A}{3}=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}=\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\Rightarrow A=\frac{9}{80}< 1\)