Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
3A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
3A - A = (\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - (\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\))
2A = 1 - \(\frac{1}{729}\) = \(\frac{728}{729}\)
A = \(\frac{728}{729}:2=\frac{364}{729}\)
a, Gọi biểu thức đó là A
Ta có :
A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
A x 3 = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{729}\)
A x 3 = \(1+A-\frac{1}{729}\)
A x 3 = \(\frac{728}{729}+A\)
A x 2 + A = \(\frac{728}{729}+A\)
A x 2 = \(\frac{728}{729}\)(bỏ A ở cả 2 vế)
A = \(\frac{728}{729}\div2=\frac{364}{729}\)
Đáp án = \(\frac{364}{729}\)
b, Phần này mình nghĩ là bạn sai đề rồi. Phải là \(\frac{45\times16-17}{45\times15+28}\)
\(\frac{46}{52}-\frac{15}{26}=\frac{46}{52}-\frac{30}{52}=\frac{16}{52}=\frac{4}{13}\)
\(\frac{15}{26}+\frac{x}{16}=\frac{46}{52}\)
\(\Leftrightarrow\frac{15}{26}+\frac{x}{16}=\frac{23}{26}\)
\(\Leftrightarrow\frac{x}{16}=\frac{23}{26}-\frac{15}{26}\)
\(\Leftrightarrow\frac{x}{16}=\frac{8}{26}=\frac{4}{13}\)
\(\Rightarrow13x=16\times4\)
\(\Rightarrow x=\frac{64}{13}\)
\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)
<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)
<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)
<=> \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{6}\) : \(\frac{5}{12}\)
,<=> \(\left(x-2\frac{1}{5}\right)\)= 2
<=. x = 2 + \(\frac{11}{5}\)
<=> x = \(\frac{21}{5}\)
Xin lỗi mk nhầm
đề là:
\(1\cdot2\cdot3\cdot4\cdot...\cdot99999999999+\left(\frac{1}{2}+\frac{2}{1}+0,5-1+3-5\right)\)
\(\frac{7}{4}-y\)x \(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
\(\frac{7}{4}-y\)x \(\frac{5}{6}=\frac{5}{6}\)
\(y\) x \(\frac{5}{6}=\frac{7}{4}-\frac{5}{6}\)
\(y\) x \(\frac{5}{6}=\frac{11}{12}\)
y == \(\frac{11}{12}:\frac{5}{6}\)
y == \(\frac{11}{10}\)
Ta có \(\frac{7}{4}-\frac{5}{6}y=\frac{5}{6}\)
\(\frac{7}{4}=\frac{5}{6}y+\frac{5}{6}\)
\(\frac{7}{4}=\frac{5}{6}\left(y+1\right)\)
\(\frac{7}{4}:\frac{5}{6}=y+1\)
\(\frac{7}{4}\cdot\frac{6}{5}=\frac{21}{10}=y+1\)
\(y=\frac{21}{10}-1=\frac{11}{10}\)
\(\frac{45+16-17}{45x15+18}\)
=\(\frac{44}{693}\)
=\(\frac{4}{63}\)