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A=1/2^2+1/3^2+...+1/23^2
=>A<1-1/2+1/2-1/3+...+1/22-1/23
=>A<22/23
\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)
Lời giải:
$\frac{1}{4}< \frac{1}{1.2}$
$\frac{1}{9}< \frac{1}{2.3}$
$\frac{1}{16}< \frac{1}{3.4}$
....
$\frac{1}{2500}< \frac{1}{49.50}$
Cộng theo vế:
$A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1$
Ta có đpcm.
Em cần làm gì để bảo tồn nề văn hóa Sa Huỳnh
Giải câu này giùm em với ạ
1)\(\dfrac{2}{9}+\dfrac{-3}{4}+\dfrac{5}{30}\)
\(=\dfrac{2.20}{9.20}+\dfrac{-3.45}{4.45}+\dfrac{5.6}{30.6}\)
\(=\dfrac{40}{180}+\dfrac{-135}{180}+\dfrac{30}{180}\)
\(=\dfrac{40+\left(-135\right)+30}{180}\)
\(=\dfrac{-65}{180}\)
\(=\dfrac{-13}{36}\)
2)\(\dfrac{-7}{12}-\dfrac{11}{18}\)
\(=\dfrac{-7.3}{12.3}-\dfrac{11.2}{18.2}\)
\(=\dfrac{-21}{36}-\dfrac{22}{36}\)
\(=\dfrac{-21-22}{36}\)
\(=\dfrac{-43}{36}\)
3)\(\dfrac{7}{8}-\dfrac{-5}{16}\)
\(=\dfrac{7.2}{8.2}-\dfrac{-5}{16}\)
\(=\dfrac{14}{16}-\dfrac{-5}{16}\)
\(=\dfrac{14-\left(-5\right)}{16}\)
\(=\dfrac{19}{16}\)
4)\(\dfrac{3}{8}-\dfrac{-9}{10}-\dfrac{5}{16}\)
\(=\dfrac{3.10}{8.10}-\dfrac{-9.8}{10.8}-\dfrac{5.5}{16.5}\)
\(=\dfrac{30}{80}-\dfrac{-72}{80}-\dfrac{25}{80}\)
\(=\dfrac{30-\left(-72\right)-25}{80}\)
\(=\dfrac{77}{80}\)
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
i don't now
mong thông cảm !
...........................
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
ta có :
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
nên \(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
\(\Rightarrow A< \frac{99}{100}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
nhiều qá lm sao nổi
b,A= \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)
\(=(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{40})+(\dfrac{1}{41}+...+1...\)
\(=(\dfrac{20}{20.21}+\dfrac{21}{21.22}+...+\dfrac{39}{39.40})+(40/...\)
\(20(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...\dfrac{1}{39.40})+40(\dfrac{1}{40}...\)
\(20(\dfrac{1}{20}-\dfrac{1}{40})+40(\dfrac{1}{40}-\dfrac{1}{60})>\dfrac{11}{15}\)
Lại có \(A<40(\dfrac{1}{20.21}+...\dfrac{1}{39.40})+60(\dfrac{1}{40.41}+...+...\)
\(=40(\dfrac{1}{20}-\dfrac{1}{40})+60(\dfrac{1}{40}-\dfrac{1}{60})<\dfrac{3}{2}\)
=> \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)
a,\( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\)
= \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+...+ \dfrac{1}{196} < \dfrac{1}{2^2-1}+ \dfrac{1}{4^2-1}+ \dfrac{1}{6^2-1}+...+ \dfrac{1}{14^2-1}\)
= \( \dfrac{1}{1.3}+ \dfrac{1}{3.5}+ \dfrac{1}{5.7}+...+ \dfrac{1}{13.15}\)
= \( \dfrac{1}{2}(1- \dfrac{1}{3}+ \dfrac{1}{3}- \dfrac{1}{5}+ \dfrac{1}{5}- \dfrac{1}{7}+ \dfrac{1}{7}-...- \dfrac{1}{13}+ \dfrac{1}{13}- \dfrac{1}{15})\)
= \( \dfrac{1}{2}(1- \dfrac{1}{15})< \dfrac{1}{2}\)
Vậy \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\) \(<\dfrac{1}{2} \)
A=1/2^2+1/3^2+...+1/23^2
=>A<1-1/2+1/2-1/3+...+1/22-1/23
=>A<22/23
tổng tử số =1+3+5+...+23
=\(\dfrac{24}{2}\)x(1+23)
=12.24
=288
tổng mẫu số=529
tổng chuỗi phân số=\(\dfrac{tongtuso}{tongmauso}\)
=\(\dfrac{288}{529}\)
=\(\dfrac{144}{265}\)
<\(\dfrac{22}{23}\)
vậy ta chứng minh được rằng:
1/4+1/9+1/16+...1/529<22/23