Tính tổng : S= 2016 + \(\frac{20\text{1}6}{\text{1}+2}\) + \(\frac{20\text{1}6}{\text{1}+2+3}\) +....+ \(\frac{20\text{1}6}{\text{1}+2+3+...+20\text{1}5}\)
\(M=\frac{\text{2 . 6 . 10 + 4 . 12 . 20 + 6 . 18 . 30 + ..... + 20 . 60 . 100}}{\text{1 . 2 . 3 + 2 . 4 . 6 + 3 . 6 . 9 + ..... + 10 . 20 . 30}}\)
Rút gọn biểu thức trên nha.
\(M=\frac{2.6.10+4.12.20+...+20.60.100}{1.2.3+2.4.6+...+10.20.30}=\frac{2.6.10.1^3+2.6.10.2^3+...+2.6.10.10^3}{1.2.3.1^3+1.2.3.2^3+...+1.2.3.10^3}\)
\(=\frac{2.6.10.\left(1^3+2^3+...+10^3\right)}{1.2.3.\left(1^3+2^3+...+10^3\right)}=\frac{2.6.10}{1.2.3}=20\)
vậy M=20
\(\frac{1\text{x}3\text{x}5+2\text{x}6\text{x}10+4\text{x}12\text{x}20+7\text{x}21\text{x}35}{1\text{x}5\text{x}7+2\text{x}10\text{x}14+4\text{x}20\text{x}28+7\text{x}35\text{x}49}\)
( hơi nhỏ xíu )
\(=\frac{1\cdot3\cdot5+2^3\cdot1\cdot3\cdot5+4^3\cdot1\cdot3\cdot5+7^3\cdot1\cdot3\cdot5}{1\cdot5\cdot7+2^3\cdot1\cdot5\cdot7+4^3\cdot1\cdot5\cdot7+7^3\cdot1\cdot5\cdot7}=\frac{1\cdot3\cdot5\cdot\left(1+2^3+4^3+7^3\right)}{1\cdot5\cdot7\cdot\left(1+2^3+4^3+7^3\right)}=\frac{3}{7}\)
Bài 4: Tính hợp lý
A=\(\frac{4}{\text{1⋅2}}+\frac{4}{\text{3⋅5}}+......+\frac{4}{\text{20⋅11⋅2013}}\)
Bài 5: So sánh với 1:
A=\(\frac{1}{\text{1⋅2}}+\frac{1}{\text{2⋅3}}+\frac{1}{\text{3⋅4}}+......+\frac{1}{\text{49⋅50}}\)
Bài 5 :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{59}\)
\(A=1-\frac{1}{50}\)
từ trên ta có : \(1-\frac{1}{50}< 1\)
\(\Rightarrow A< 1\)
chứng tỏ rằng: A= \(\frac{\text{1}}{2\text{1}}\) +\(\frac{2}{3\text{1}}\) +...+ \(\frac{20\text{1}3}{20\text{1}4}\) < 1
sửa lại đề : Chứng tỏ rằng : A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}< 1\)
bài làm
A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}\)
A = \(\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2014-1}{2014!}\)
A = \(1-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2014}{2014!}-\frac{1}{2014!}\)
A = \(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2013!}-\frac{1}{2014!}\)
A = \(1-\frac{1}{2014!}< 1\)
\(\text{Cho }P=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{399}{400}\text{ Chứng minh }P< \frac{1}{20}\)
\(P=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}......\frac{399}{400}\)
\(P=\frac{1.3.4.5....399}{2.4.5.6.....400}\)
\(P=\frac{1.3}{2.400}\)
\(P=\frac{3}{800}\)
Vì \(\frac{3}{800}< \frac{40}{800}\)
\(\Rightarrow P< \frac{40}{800}\)
\(\Rightarrow P< \frac{1}{20}\left(đpcm\right)\)
Ta co:
\(P=\frac{1}{2}.\frac{3.4.5...399}{4.5.6...400}\)
\(\Leftrightarrow P=\frac{1}{2}.\frac{3}{400}=\frac{3}{800}< \frac{3}{600}=\frac{1}{20}\)
\(\Rightarrow P< \frac{1}{20}\left(dpcm\right).\)
\(P=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}......\frac{399}{400}\)
\(P=\frac{1.3.4.5.....399}{2.4.5.6....400}\)
\(P=\frac{1.3}{2.400}\)
\(P=\frac{3}{800}\)
\(V\text{ì}\frac{3}{800}< \frac{40}{800}\)
\(\Rightarrow P< \frac{40}{800}\)
\(\Rightarrow P< \frac{1}{20}\left(\text{đ}pcm\right)\)
Rút gọn phân số sau:
\(M=\frac{\text{2 . 6 . 10 + 4 . 12 . 20 + 6 . 18 . 30 + ..... + 20 . 60 . 100 }}{\text{1 . 2 . 3 + 2 . 4 . 6 + 3 . 6 . 9 + ..... + 10 . 20 . 30}}\)
\(M=\frac{2.6.10+4.12.20+6.18.30+...+20.60.100}{1.2.3+2.4.6+3.6.9+...+10.20.30}\)
\(=\frac{2.6.10.\left(1+2+3+...+10\right)}{1.2.3.\left(1+2+3+...+10\right)}\)
\(=20\)
\(\frac{2017}{1\times2\text{×}3}+\frac{2017}{2\text{×}3\text{×}4}+\frac{2017}{3\text{×}4\text{×}5}+..+\frac{2017}{19\text{×}20\text{×}21}\)
\(\frac{2017}{1.2.3}+\frac{2017}{2.3.4}+\frac{2017}{3.4.5}+...+\frac{2017}{19.20.21}\)
\(=2017\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\right)\)
\(=2017.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}\right)\)
\(=2017.\left(1-\frac{1}{2}-\frac{1}{3}-\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)-...-\left(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\right)\right)\)
\(=2017.\left(1+\frac{1}{21}\right)\)phá ngoặc trước dấu trừ đổi dấu,rút gọn:
\(=2017.\frac{20}{21}=\frac{40340}{21}\)
\(\frac{\text{2/36 - 1/15 - 2/153}}{\text{1/34 + 3/20 - 3/26 }}:\frac{\text{1 + 2/71 - 5/121}}{\text{65/121 - 26/71 - 13 }}\)
bai 1 \(\frac{-3}{\text{2}}+\frac{5}{7}+\frac{-31}{14}< hoac=\text{x}< \frac{1}{\text{2}}+\frac{1}{3}+\frac{1}{6}\)\(\frac{1}{6}\)
bai 2 \(\frac{\text{x}+4}{\text{x}-\text{2}}+\frac{\text{2}\text{x}-5}{\text{x}-\text{2}}\)la so nguyen
Bài 1 mk ko hiểu đề cho lắm
Bài 2 :
Đặt \(A=\frac{x+4}{x-2}+\frac{2x-5}{x-2}\)
Ta có :
\(\frac{x+4}{x-2}+\frac{2x-5}{x-2}=\frac{x+4+2x-5}{x-2}=\frac{3x-1}{x-2}=\frac{3x-6+5}{x-2}=\frac{3\left(x-2\right)}{x-2}+\frac{5}{x-2}=3+\frac{5}{x-2}\)
Để \(A\) là số nguyên thì \(\frac{5}{x-2}\) phải là số nguyên \(\Rightarrow\) \(5⋮\left(x-2\right)\) \(\Rightarrow\) \(\left(x-2\right)\inƯ\left(5\right)\)
Mà \(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)
Do đó :
\(x-2\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
\(x\) | \(3\) | \(1\) | \(7\) | \(-3\) |
Vậy \(x\in\left\{-3;1;3;7\right\}\) thì A là số nguyên
Chúc bạn học tốt ~
bai 1: Viết tập hợp A các số nguyên x biết:
cau hoi cua phung minh quan