bạnh ơi nếu rảnh thì lm toán nha 

Ai làm được không?

easy but i don't know

Ta có \(H=\frac{7}{3}+\frac{13}{3^2}+...+\frac{605}{3^{100}}\)

\(\Leftrightarrow3H=7+\frac{13}{3}+...+\frac{605}{3^{99}}\)

\(\Rightarrow2H=7+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)

\(\Leftrightarrow2H=7+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)

Mà \(6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)=3-\frac{1}{3^{99}}\)

\(\Rightarrow2H=7+3-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)

\(\Leftrightarrow2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)

Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)

\(\Rightarrow2H< 10\)

\(\Leftrightarrow H< 5\left(1\right)\)

Ta có \(2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)

Mà\(\frac{1}{3^{97}}+\frac{605}{3^{98}}< 22\)

hay\(\frac{1}{3^{99}}+\frac{605}{3^{98}}< \frac{22}{9}\)

\(\Rightarrow2H>10-\frac{22}{9}=\frac{68}{9}=2\cdot\left(3+\frac{7}{9}\right)\)

\(\Rightarrow H>3+\frac{7}{9}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Rightarrowđpcm\)

Sai r

Tham khảo nha bạn :

Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến

a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)

\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)

Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)

\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)

\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)

\(\Rightarrow F< \frac{3}{2}\)

\(\Rightarrow2A< 4+\frac{3}{2}\)

\(\Rightarrow2A< \frac{11}{2}\)

\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)

2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)

\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)

\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)

\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)

Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)

\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)

\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )

\(\Rightarrow2D< 6\)

\(\Rightarrow D< 3\)

\(\Rightarrow2B< 11+3\)

\(\Rightarrow2B< 14\)

\(\Rightarrow B< 7\left(đpcm\right)\)

Bài 1:
Ta có:

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Mà \(\frac{99}{100}< 1\)

\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)

Có phải ở sách NCPT ko bn

\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)

\(A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)

\(A=1-\frac{1}{100}=\frac{99}{100}< 1\)

\(\Rightarrow A< 1\text{(đpcm) }\)