Khó dữ vậy!!!!

Đợi tí , mạng chậm

Có : \(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow2A< 1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

Có: \(6A< 3+1+\frac{1}{3}+...+\frac{1}{3^{98}}\)

\(6A-2A< 3-\frac{1}{3^{99}}< 3\)

\(\Rightarrow4A< 3\Rightarrow A< \frac{3}{4}\)(đpcm)

a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)

\(\Leftrightarrow\left(\frac{1}{5}\right)^{3n-1}=\left(\frac{1}{5}\right)^2\)

\(\Leftrightarrow3n-1=2\)

\(\Leftrightarrow3n=3\)

\(\Leftrightarrow n=1\)

b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)

\(\Leftrightarrow\left(\frac{4}{7}\right)^{n+2}=\left(\frac{4}{7}\right)^{-1}\)

\(\Leftrightarrow n+2=-1\)

\(\Leftrightarrow n=-3\)

c)\(\left(\frac{2}{3}\right)^{-n+1}=\frac{3^3}{2^3}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{3}{2}\right)^3\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{2}{3}\right)^{-3}\)

\(\Leftrightarrow-n+1=-3\)

\(\Leftrightarrow n=-4\)

c)\(\left(0,7\right)^{3n+1}=10^3:7^3\)

\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{10}{7}\right)^3\)

\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{7}{10}\right)^{-3}\)

\(\Leftrightarrow3n+1=-3\)

\(\Leftrightarrow3n=-4\)

\(\Leftrightarrow n=-\frac{4}{3}\)

Bạn tham khảo ở link này nhé :

Câu hỏi của Tăng Minh Châu - Toán lớp 6 | Học trực tuyến

\(a,15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\\ =15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\\ =7\frac{3}{13}-3\frac{4}{7}\\ =\frac{94}{13}-\frac{25}{7}\\ =\frac{94\cdot7-25\cdot13}{13\cdot7}\\ =\frac{333}{91}\)

\(b,\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\\ =7\frac{4}{9}-3\frac{4}{9}+4\frac{7}{11}\\ =4\frac{4}{9}+4\frac{7}{11}\\ =\frac{40}{9}+\frac{51}{11}\\ =\frac{40\cdot11+51\cdot9}{11\cdot9}\\ =\frac{899}{99}\)

\(c,=-\frac{7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\\ =-\frac{7}{9}\cdot1+\frac{52}{9}\\ =\frac{-7+52}{9}\\ =\frac{45}{9}=5\)

\(d,=\frac{50}{100}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{75}{100}\\ =\frac{50\cdot2\cdot2\cdot2\cdot5\cdot7\cdot25\cdot3}{50\cdot2\cdot3\cdot5\cdot7\cdot25\cdot2\cdot2}=1\)

\(e,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\\ =1-\frac{1}{43}\\ =\frac{42}{43}\)

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)