\(4\frac{5}{6}\):\(5\frac{6}{7}\)=?
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(\frac{-5}{6}\right)-\frac{6}{7}-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(-\frac{5}{6}\right)-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{7}{8}+\frac{6}{7}\)
\(=\frac{7}{8}+\frac{6}{7}=\frac{49}{56}+\frac{48}{56}=\frac{49+48}{56}=\frac{97}{56}\)
So sánh:\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}+\frac{\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{4}{3}}{\frac{2}{1}}}\) và\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}+\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{5}{6}}{\frac{7}{8}}+\frac{\frac{4}{3}}{\frac{2}{1}}}\)và \(\frac{\frac{\frac{1}{2}+\frac{8}{7}}{\frac{3}{4}+\frac{6}{5}}}{\frac{\frac{5}{6}+\frac{4}{3}}{\frac{7}{8}+\frac{2}{1}}}\)và\(\frac{\frac{\frac{1+8}{2+7}}{\frac{3+6}{4+5}}}{\frac{5+4}{\frac{6+3}{2+1}}}\)
Tính nhanh: \(\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\frac{6}{4!+5!+6!}+\frac{7}{5!+6!+7!}+\frac{8}{6!+7!+8!}\)
Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)
81.\(\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right)\))
Ai đúng tick giải theo cách lớp 6 nha
Ta có \(81.\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{3}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right)\)
\(=81.\left(\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right)\)
\(=81.\left(\frac{12}{4}:\frac{5}{6}\right)\)
\(=81.\frac{18}{5}\)
\(=291,6\)
\(81\left(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{159}+\frac{6}{91}}\right)\)
\(=81\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{2}{169}+\frac{1}{91}\right)}\right)\)
\(=81\left(3\div\frac{5}{6}\right)\)
\(=81.\frac{18}{5}\)
\(=\frac{1458}{5}\)
81.[12(1-1/7-1/289-1/85)/4(1-1/7-1/289-1/85):5(1+1/13+1/169+1/91)/6(1/13+1/169+1/91)
=81.[12/4:5/6]=81.[12/4.6/5]=81.18/5=1458/5
So sánh:
A=\(\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\) và B=\(\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\)
cách này mình tự nghĩ
\(\hept{\begin{cases}A=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\\B=\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\end{cases}}\)
\(\Rightarrow A-B=\left(\frac{4}{7}-\frac{4}{7}\right)+\left(\frac{5}{7^3}-\frac{5}{7^3}\right)+\left(5-5\right)+\left(\frac{3}{7^2}-\frac{6}{7^2}\right)+\left(\frac{6}{7^4}-\frac{5}{7^4}\right)\)
\(\Rightarrow A-B=-\frac{3}{7^2}+\frac{1}{7^4}\)
\(\Rightarrow A-B=\frac{-3\times7^2}{7^4}+\frac{1}{7^4}\)
mà \(-3\times7^2< 1\Rightarrow\frac{1}{7^4}>\frac{-3\times7^2}{7^4}\Rightarrow B>A\)
Tính hợp lý
\(\frac{\frac{5}{3}+\frac{5}{8}-\frac{5}{7}}{\frac{-4}{3}-\frac{-4}{8}+\frac{4}{7}}:\frac{\frac{2}{3}-\frac{1}{6}+\frac{6}{7}}{\frac{-1}{3}+\frac{1}{6}-\frac{1}{7}}\)
\(=\frac{5\left(\frac{1}{3}+\frac{1}{8}-\frac{1}{7}\right)}{-4\left(\frac{1}{3}+\frac{1}{8}-\frac{1}{7}\right)}:\frac{2\left(\frac{1}{3}-\frac{1}{12}+\frac{3}{7}\right)}{ }\)
MÃu thứ hai sao ý
Thực hiện phép tính:
\(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{3\left(12-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}\right)}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{79.2.1001001}{79.9.1001001}\)
\(B=81.\left[3.\frac{6}{5}\right].\frac{2}{9}\)
\(B=\frac{9.9.3.6.2}{5.9}\)
\(B=\frac{9.3.6.2}{5}\)
\(B=\frac{324}{5}\)
Tick cho minh nha Quang Hai Duong tick minh may man ca nam
Bài 1: Tính(hợp lý nếu có thể) a) \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\) b) \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\) c) \(\frac{-3}{5}.\frac{5}{7}+\frac{-3}{5}.\frac{3}{7}+\frac{-3}{5}.\frac{6}{7}\) d) \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}\)
Bài 1:
a) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=4\frac{5}{7}-1\frac{3}{4}\)
\(=\frac{33}{7}-\frac{7}{4}\)
\(=\frac{132}{28}-\frac{49}{28}=\frac{83}{28}\)
b) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)
\(=4\frac{5}{9}-2\frac{3}{4}\)
\(=\frac{41}{9}-\frac{11}{4}\)
\(=\frac{164}{36}-\frac{99}{36}=\frac{65}{36}\)
c) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\cdot\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=\frac{-3}{5}\cdot2=-\frac{6}{5}\)
d) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
1. So sánh A và B:
\(A=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\); \(B=\frac{5}{7^3}+\frac{6}{7^2}+\frac{5}{7^4}+\frac{4}{7}+5\)
\(\frac{14+\frac{7}{15}+\frac{7}{4}}{2+\frac{1}{15}+\frac{1}{4}}:\frac{5+\frac{5}{17}+\frac{5}{13}}{6+\frac{6}{17}+\frac{6}{13}}+\frac{5858}{5050}\)
Giúp mình với
Tính tất cả ra thì được:
\(=\frac{\frac{973}{60}}{\frac{139}{60}}:\frac{\frac{1255}{221}}{\frac{1506}{221}}+\frac{5858}{5050}\)
\(=\frac{\frac{139}{60}}{\frac{973}{60}}.\frac{\frac{1506}{221}}{\frac{1255}{221}}+\frac{5858}{5050}\)
Tính tử và mẫu dần rồi ra ( phần này dễ mà )
Ta được: ( mình chỉ lấy 2 chữ số phần thập phân thôi )
\(=\frac{1578}{9209}+\frac{5858}{5050}\)
= 133/100
End