\(1\frac{1}{9}\cdot1\frac{1}{10}\cdot1\frac{1}{11}\cdot...\cdot1\frac{1}{2011}\)
dấu chấm ở giữa kia là dấu nhân
rút gọn phân số \(1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot1\frac{1}{24}\cdot...\cdot1\frac{1}{360}\)
1<1/3+1/8+1/15+1/24+....+1/360>
KO BIẾT ĐÚNG HAY KO NHÉ BẠN
\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}...1\frac{1}{360}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{361}{360}\)
\(=\frac{2^2}{3}.\frac{3^2}{8}.\frac{4^2}{15}.\frac{5^2}{24}...\frac{19^2}{360}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}...\frac{19.19}{18.20}\)
\(=\left(\frac{2.3.4.5...19}{1.2.3.4...18}\right).\left(\frac{2.3.4.5...19}{3.4.5.6...20}\right)\)
\(=19.\frac{1}{10}\)
\(=\frac{19}{10}\)
Rút gọn:
\(A=1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot1\frac{1}{24}\cdot...\cdot1\frac{1}{360}\)
\(A=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.....1\frac{1}{360}\)
\(A=1+\left(\frac{1}{3}.\frac{1}{8}.\frac{1}{15}.\frac{1}{24}.....\frac{1}{360}\right)\)
Nếu đúng thì tk nha
\(1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot..........\cdot1\frac{1}{9800}\)tinh nhanh
tính:
\(1\frac{1}{38}\cdot1\frac{1}{39}\cdot1\frac{1}{40}\cdot\)...(cho đến)...\(\cdot1\frac{1}{2013}\)
Bg
\(1\frac{1}{38}.1\frac{1}{39}.1\frac{1}{40}\)...\(.1\frac{1}{2013}\)
= \(\frac{39}{38}.\frac{40}{39}.\frac{41}{40}.\)...\(.\frac{2014}{2013}\)
= \(\frac{39.40.41.....2014}{38.39.40......2013}\)(39 trên, 39 dưới, 40 trên, 40 dưới,... 2013 trên, 2013 dưới, chịt tiêu hết)
= \(\frac{2014}{38}\)
= 53
cảm ơn bẹn nhe
\(1\frac{1}{38}.1\frac{1}{39}.1\frac{1}{40}...1\frac{1}{2013}=\frac{39}{38}.\frac{40}{39}.\frac{41}{40}...\frac{2014}{2013}=\frac{2014}{38}=53\)
Do thấy họ của bạn giống họ của tui nên giúp
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2011}}{\frac{1}{2011\cdot1}+\frac{1}{3\cdot2009}+...+\frac{1}{2011\cdot1}}\)
\(M=1\frac{2}{3}\cdot1\frac{2}{5}\cdot1\frac{2}{7}\cdot...\cdot1\frac{2}{97}\)
1) Tính nhanh:
P=\(1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot1\frac{1}{24}\cdot1\frac{1}{35}\cdot1\frac{1}{48}\cdot1\frac{1}{63}\cdot1\frac{1}{80}\)
2) So sánh:
A=\(\frac{100^{10}+1}{100^{10}-1}\) và B=\(\frac{100^{10}-1}{100^{10}-3}\)
3) So sánh A và B biết:
A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
B=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
3)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{49}{50}\)
\(=\frac{49}{50}\)
⇒ A < 1 (1)
\(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow B>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{10}+\frac{90}{100}=1\)
⇒ B > 1 (2)
từ (1) và (2) ⇒ A<1<B
vậy A < B
\(1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot...\) có 98 thừa số
Ta có \(1\frac{1}{3}=\frac{2^2}{3};1\frac{1}{8}=\frac{3^2}{8};.....\)
Nên thừa số thứ 98 là : \(1\frac{1}{9800}=\frac{99^2}{9800}\)
Ta có \(\frac{2^2}{3}.\frac{3^2}{8}......\frac{99^2}{9800}=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}=\frac{2.2.3.3.....99.99}{1.3.2.4....98.100}\)
\(=\frac{\left(2.3.4...99\right).\left(2.3.4....99\right)}{\left(1.2.3....98\right).\left(3.4.5...100\right)}=\frac{99.2}{1.100}=\frac{198}{100}=\frac{99}{50}\)
tìm x
\(\frac{2017}{2017\cdot\left(x\cdot1\right)}\)=21
dấu chấm là dấu nhân nha ghi đầy đủ cách giải
ta có x.1=21(vì 2017:2017=1 nên triệt tiêu)
x =21:1=21
\(\frac{2017}{2017\cdot\left(x\cdot1\right)}=21\)
\(\Leftrightarrow\frac{1}{x}=21\)
\(\Leftrightarrow x=\frac{1}{21}\)
\(\frac{2017}{2017.\left(x.1\right)}=21\)
\(\frac{1}{x}=21\)
\(x=\frac{1}{21}\)