Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3: \(=20-12-8+12=20-8=12\)
5: \(=-18-42-21-35=-116\)
3: \(=-15+18-12+8=-27+26=-1\)
2: \(=-12+21-15+10=9-5=4\)
a) \(\frac{7}{2}-\frac{14}{3}+\left(\frac{3}{4}-\frac{7}{3}\right)-\left(\frac{5}{6}-\frac{7}{4}\right)+\frac{11}{2}-3\)
\(=\frac{7}{2}-\frac{14}{3}+\frac{3}{4}-\frac{7}{3}-\frac{5}{6}+\frac{7}{4}+\frac{11}{2}-3\)
\(=\left(\frac{7}{2}+\frac{11}{2}\right)-\left(\frac{14}{3}+\frac{7}{3}\right)-3+\left(\frac{3}{4}+\frac{7}{4}\right)-\frac{5}{6}\)
\(=9-7-3+(\frac{5}{2}-\frac{5}{6})\)
\(=-1+\frac{5}{3}\)
\(=\frac{2}{3}\)
b) \(\frac{7}{3}-\frac{7}{5}+\frac{11}{10}-\left(\frac{2}{5}-\frac{5}{6}\right)+\frac{4}{15}-\frac{4}{3}\)
\(=\left(\frac{7}{3}-\frac{4}{3}\right)-\left(\frac{7}{5}+\frac{2}{5}\right)+(\frac{11}{10}+\frac{5}{6}+\frac{4}{15})\)
\(=1-\frac{9}{5}+\frac{11}{5}\)
\(=1-\left(\frac{9}{5}-\frac{11}{5}\right)\)
\(=1-\left(\frac{-2}{5}\right)\)
\(=1\frac{2}{5}\)
......................?
mik ko biết
mong bn thông cảm
nha ................
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
\(6A=7^{2008}-1\)
\(A=\frac{7^{2008}-1}{6}\)
Tương tự, \(B=\frac{4^{101}-1}{3},C=\frac{3^{101}-1}{2}\).
\(D=7+7^3+7^5+7^7+...+7^{99}\)
\(7^2.D=7^3+7^5+7^7+7^9+...+7^{101}\)
\(\left(7^2-1\right)D=\left(7^3+7^5+7^7+7^9+...+7^{101}\right)-\left(7+7^3+7^5+7^7+...+7^{99}\right)\)
\(48D=7^{101}-7\)
\(D=\frac{7^{101}-7}{48}\)
Tương tự, \(E=\frac{2^{9011}-2}{3}\)
1,
3/5 - (3/5+1/3)
= 3/5 - 3/5 - 1/3
= 0 - 1/3
= -1/3
2.
3/7 x 4/5 + 3/7 x 1/5 - 2/7
= 3/7.(4/5+1/5) - 2/7
= 3/7 . 1 - 2/7
= 3/7-2/7 = 1/7
1) 3/5 - (3/5 + 1/3)
= 3/5 - 3/5 +1/3
= 0+1/3
=1/3
2) 3/7x4/5+3/7.1/5-2/7
=3/7.(4/5+1/5)-2/7
=3/7.1-2/7
=3/7-2/7
=1/7
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
