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Ta có: A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2013.2015}\)
\(\Leftrightarrow2A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}\right)\)
\(\Leftrightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2013}+\dfrac{1}{2013}-\dfrac{1}{2015}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{2015}=\dfrac{2012}{6045}\)
\(\Leftrightarrow A=\dfrac{1006}{6045}\)
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{2013.2015}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}+\dfrac{1}{2015}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{2015}\)
2A=\(\dfrac{2014}{2015}\)
A=\(\dfrac{1007}{2015}\)
Khi gặp bài này, bn nên tách 1 phân số ra thành hiệu của 2 phân số.
Gọi d=ƯCLN(14n+3;21n+5)
=>42n+9-42n-10 chia hết cho d
=>-1 chia hết cho d
=>d=1
=>PSTG
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
Ta có : \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
mà \(2019< 2020\)nên P < 1 ( đpcm )
\(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\)
\(P=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\)
\(P=1-\dfrac{1}{2021}\)
\(P=\dfrac{2020}{2021}\)
Vì \(\dfrac{2020}{2021}< 1\) ⇒ \(P< 1\) ( điều phải chứng minh )
Giả sử UCLN(14n+3;21n+5)=d
14n+3 chia hết cho d nên 42n+9 chia hết cho d
21n+5 chia hết cho d nên 42n+10 chia hết cho d
vay 1 chia hết cho d, d=1
Vậy phân số tối giản
Giải:
Gọi ƯC(14n+3;21n+5)=d
⇒14n+3 ⋮ d ⇒3.(14n+3) ⋮ d ⇒42n+9 ⋮ d
21n+5 ⋮ d 2.(21n+5) ⋮ d 42n+10 ⋮ d
⇒(42n+10)-(42n+9) ⋮ d
⇒ 1 ⋮ d
⇒d=1
Vậy 14n+3/21n+5 là phân số tối giản.
Chúc bạn học tốt!
a: \(A=\dfrac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x}{x^2+x+1}\)
a/(b+c) + b/(a+c) + c/(a+b) = a^2/(ab+ac) + b^2/(ba+bc) + c^2/(ac+bc) >=
(a+b+c)^2/(2.(ab+bc+ac) (buhihacopxki dạng phân thức)
>= (3.(ab+bc+ac)/(2(ab+bc+ac) =3/2
a^2/(b^2+c^2) + b^2/(a^2+c^2) + c^2/(a^2+b^2) >= (a+b+c)^2/(2.(a^2+b^2+c^2) (buhihacopxki dạng phân thức)
>= 3(a^2+b^2+c^2) / 2(a^2+b^2+c^2) >=3/2
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}-\dfrac{3}{2}\ge0\)
\(\Leftrightarrow\left(\dfrac{a}{b+c}-\dfrac{1}{2}\right)+\left(\dfrac{b}{c+a}-\dfrac{1}{2}\right)+\left(\dfrac{c}{a+b}-\dfrac{1}{2}\right)\ge0\)
\(\Leftrightarrow\left(\dfrac{2a-b-c}{2\left(b+c\right)}\right)+\left(\dfrac{2b-a-c}{2\left(a+c\right)}\right)+\left(\dfrac{2c-a-b}{2\left(a+b\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b+a-c}{2\left(b+c\right)}+\dfrac{b-a+b-c}{2\left(a+c\right)}+\dfrac{c-a+c-b}{2\left(a+b\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(b+c\right)}+\dfrac{a-c}{2\left(b+c\right)}+\dfrac{b-a}{2\left(a+c\right)}+\dfrac{b-c}{2\left(a+c\right)}+\dfrac{c-a}{2\left(a+b\right)}+\dfrac{c-b}{2\left(a+b\right)}\ge0\)\(\Leftrightarrow\left(a-b\right)\left[\dfrac{1}{2\left(b+c\right)}-\dfrac{1}{2\left(a+c\right)}\right]+\left(a-c\right)\left[\dfrac{1}{2\left(b+c\right)}-\dfrac{1}{2\left(a+b\right)}\right]+\left(b-c\right)\left[\dfrac{1}{2\left(a+c\right)}-\dfrac{1}{2\left(a+b\right)}\right]\ge0\)
ta có: a,b,c là 3 số dương bất kì nên ta giả sử \(a\ge b\ge c\)
\(\Rightarrow a+c\ge b+c\)
\(\Leftrightarrow2\left(a+c\right)\ge2\left(b+c\right)\)
\(\Leftrightarrow\dfrac{1}{2\left(a+c\right)}\le\dfrac{1}{2\left(b+c\right)}\)
\(\Leftrightarrow\dfrac{1}{2\left(a+c\right)}-\dfrac{1}{2\left(b+c\right)}\ge0\)
Mà \(a\ge b\Rightarrow a-b\ge0\)
\(\Rightarrow\left(a-b\right)\left[\dfrac{1}{2\left(b+c\right)}-\dfrac{1}{2\left(a+c\right)}\right]\ge0\left(1\right)\)
Chứng minh tương tự, ta có:
\(\left(a-c\right)\left[\dfrac{1}{2\left(b+c\right)}-\dfrac{1}{2\left(a+b\right)}\right]\ge0\left(2\right)\)
\(\left(b-c\right)\left[\dfrac{1}{2\left(a+c\right)}-\dfrac{1}{2\left(a+b\right)}\right]\ge0\left(3\right)\)
Cộng từng vế (1);(2);(3) \(\Rightarrow\) luôn đúng
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
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\(\dfrac{1009}{2019}\)
b) Gọi ƯCLN(14n+3,21n+5) là d
14n+3⋮d ⇒42n+9⋮d
21n+5⋮d ⇒42n+10⋮d
(42n+10)-(42n+9)⋮d
1⋮d ⇒ƯCLN(14n+3,21n+5)=1
Vậy \(\dfrac{14n+3}{21n+5}\) là Ps tối giản