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Với a,b,c dương, ta có:
a/a+b > a/a+b+c
b/b+c > b/a+b+c
c/c+a > c/a+b+c
=> A > a/a+b+c + b/a+b+c + c/a+b+c => A>1. (1)
Ta lại có
A = a/a+b + b/b+c + c/c+a
= a+b-b/a+b + b+c-c/b+c + c+a-a/c+a
= 1-b/a+b + 1-c/b+c + 1-a/c+a
= 3-(b/a+b + c/b+c + a/c+a) = 3-B
Tương tự phần chứng minh trên, ta có
b/a+b > b/a+b+c
c/b+c > c/a+b+c
a/a+c > a/a+b+c
=> B > b/a+b+c + c/a+b+c + a/a+b+c => B>1
mà A = 3-B
=> A < 2 (2)
Từ (1) và (2) => 1<A<2
Mà không có số tự nhiên nào ở giữa 1 và 2 => A không là số tự nhiên
\(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + .....+ \(\dfrac{1}{10.11.12}\)
= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\) - \(\dfrac{1}{3.4}\) +....+ \(\dfrac{1}{10.11}\) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{1.2}\) + (- \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\))+.......+ ( \(-\dfrac{1}{10.11}\) + \(\dfrac{1}{10.11}\)) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{11.12}\) =\(\dfrac{1}{2}\) - \(\dfrac{1}{132}\) =\(\dfrac{66}{132}\)-\(\dfrac{1}{132}\) =\(\dfrac{65}{132}\) Vì \(\dfrac{33}{132}\) = \(\dfrac{1}{4}\) nên \(\dfrac{65}{132}\) > \(\dfrac{1}{4}\)
a: Gọi phân số cần tìm có dạng là \(\dfrac{a}{b}\left(b\ne0\right)\)
Theo đề, ta có: \(\dfrac{1}{3}< \dfrac{a}{b}< \dfrac{1}{2}\)
=>\(0,\left(3\right)< \dfrac{a}{b}< 0,5\)
=>\(\dfrac{a}{b}=0,4;\dfrac{a}{b}=0,42\)
=>\(\dfrac{a}{b}=\dfrac{2}{5};\dfrac{a}{b}=\dfrac{21}{25}\)
Vậy: Hai phân số cần tìm là \(\dfrac{2}{5};\dfrac{21}{25}\)
b: a/b<1
=>a<b
=>\(a\cdot c< b\cdot c\)
=>\(a\cdot c+ab< b\cdot c+ab\)
=>\(a\left(c+b\right)< b\left(a+c\right)\)
=>\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
a) S=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2017.2019}\)
2S=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
2S=\(1-\dfrac{1}{2019}\)
2S=\(\dfrac{2018}{2019}\)
S\(\dfrac{1009}{2019}\)
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
Lời giải:
$(\frac{2023a}{2024c})^3=(\frac{2024b}{2025a})^3=(\frac{2025c}{2023b})^3=\frac{2023a}{2024b}.\frac{2024b}{2025a}.\frac{2025c}{2023b}=1$
$\Rightarrow \frac{2023a}{2024c}=\frac{2024b}{2025a}=\frac{2025c}{2023b}=1$
$\Rightarrow 2023a=2024c; 2024b=2025a; 2025c=2023b$
Do đó:
$\frac{2023a}{506c}+\frac{2024b}{675a}+\frac{2025c}{289b}=\frac{2024c}{506c}+\frac{2025a}{675a}+\frac{2023b}{289b}$
$=4+3+7=14$