Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
Kiểm tra lại đề xem thừa số cuối có đúng quy luật của dãy không.
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
Đặt A=\(\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
A=\(\dfrac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}\)
A=\(\dfrac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
A=2.\(5^2\)
A=2.25
A=50
\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1-\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}=\dfrac{1}{2017}\)
Giải:
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right).\left(1-\dfrac{1}{2017}\right)\)
\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}.\dfrac{2016}{2017}\)
\(\Leftrightarrow A=\dfrac{1.2...201.2016}{2.3...2016.2017}\)
\(\Leftrightarrow A=\dfrac{1.2.3...2015.2016}{2017.2.3...2015.2016.}\)
Rút gọ cả tử và mẫu với 2.3...2015.2016, ta được:
\(A=\dfrac{1}{2017}\)
Vậy \(A=\dfrac{1}{2017}\).
Chúc bạn học tốt!
Bài 2:
a: \(=44\cdot82-400+18\cdot44\)
\(=44\cdot100-400=4400-400=4000\)
b: \(=6^2:\left\{780:\left[390-125\cdot49+65\right]\right\}\)
\(=36:\left\{780:\left[-5670\right]\right\}\)
\(=36:\dfrac{-26}{189}=\dfrac{-3402}{13}\)