Tính:\(A=3\dfrac{1}{417}.\dfrac{1}{762}-\dfrac{1}{139}+\dfrac{761}{762}-\dfrac{4}{417.762}+\dfrac{5}{139}\)
LƯU Ý:\(3\dfrac{1}{417}\)là hỗn số
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LS
2 tháng 3 2019
\(M=\left(\dfrac{1252}{417.762}-\dfrac{4}{417.762}\right)-\left(\dfrac{1}{139}+\dfrac{5}{139}\right).\dfrac{3809}{762}\)
\(M=\left(\dfrac{1252-4}{317754}\right)-\dfrac{6}{139}.\dfrac{3809}{762}\)
\(M=\left(\dfrac{1248}{317754}\right)-\dfrac{22854}{105918}\)
\(M=\dfrac{208}{52959}-\dfrac{3809}{17653}\)
\(M=\dfrac{3671824}{934885227}-\dfrac{201720831}{934885227}\)
\(M=\dfrac{-198049007}{934885227}\)
5 tháng 4 2017
Bài 2:
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=...=\dfrac{a_9-9}{1}=\dfrac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}=\dfrac{\left(a_1+a_2+...+a_9\right)-\left(1+2+...+9\right)}{9+8+...+1}\)
\(=\dfrac{90-45}{45}=1\)
+) \(\dfrac{a_1-1}{9}=1\Rightarrow a_1=10\)
+) \(\dfrac{a_2-2}{8}=1\Rightarrow a_2=10\)
...
+) \(\dfrac{a_9-9}{1}=1\Rightarrow a_9=10\)
Vậy \(a_1=a_2=...=a_9=10\)
10 tháng 12 2022
Câu 2
(a+3)(b-4)-(a-3)(b+4)=0
=>ab-4a+3b-12-ab-4a+3b+12=0
=>-8a=-6b
=>a/b=3/4
=>a/3=b/4
22 tháng 11 2022
Đặt 761=a; 139=b
\(M=\left(3+\dfrac{1}{3b}\right)\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\left(4+\dfrac{a}{a+1}\right)-\dfrac{4}{\left(a+1\right)\cdot3b}+\dfrac{5}{b}\)
\(=\dfrac{9b+1}{3b}\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\dfrac{4a+4}{a+1}-\dfrac{4}{3b\left(a+1\right)}+\dfrac{5}{b}\)
\(=\dfrac{9b+1-3\left(4a+4\right)-4+15\left(a+1\right)}{3b\left(a+1\right)}\)
\(=\dfrac{9b+1-12a-12-4+15a+15}{3b\left(a+1\right)}\)
\(=\dfrac{3a+9b}{3b\left(a+1\right)}=\dfrac{3\left(a+3b\right)}{3b\left(a+1\right)}=\dfrac{a+3b}{b\left(a+1\right)}\)
\(=\dfrac{761+3\cdot139}{139\left(761+1\right)}=\dfrac{589}{52959}\)
10 tháng 4 2019
\(D=\left(3+\frac{1}{417}\right).\frac{1}{762}-\frac{1}{139}\left(4+\frac{761}{762}\right)-\frac{4}{417.762}+\frac{5}{139}\)
=\(\frac{3}{762}+\frac{1}{417.762}-\frac{4}{139}-\frac{761}{139.762}-\frac{4}{417.762}+\frac{5}{139}\)
=\(\frac{3}{762}-\frac{1}{139.762}+\frac{1}{139}-\frac{761}{139.762}=\frac{3}{762}+\frac{1}{139}\left(-\frac{1}{762}+1\right)-\frac{761}{139.762}=\)
\(\frac{3}{762}+\frac{761}{139.762}-\frac{761}{139.762}=\frac{3}{762}\)
\(A=3\dfrac{1}{417}.\dfrac{1}{762}-\dfrac{1}{139}+\dfrac{761}{762}-\dfrac{4}{417.762}+\dfrac{5}{139}\)
\(=\left(\dfrac{1252}{417.762}-\dfrac{4}{417.762}\right)+\left(-\dfrac{1}{139}+\dfrac{5}{139}\right)+\dfrac{761}{762}\)
\(=\dfrac{1248}{417.762}+\dfrac{4}{139}+\dfrac{761}{762}=\dfrac{1248}{417.672}+\dfrac{12.762}{417.762}+\dfrac{761.417}{417.762}\)
\(=\dfrac{327729}{317754}\)