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Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
A = 7/4 . (3333/1212 + 3333/2020 + 3333/3030 + 3333/4242)
A = 7/4 . (11/4 + 33/20 + 11/10 + 11/14)
A = 7/4 . 44/7
A = 11
Chúc bạn học tốt
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{42.101}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{6}\right)\)
\(A=\frac{7}{4}.33.\frac{1}{6}\)
\(A=\frac{7.33}{4.6}\)
\(A=\frac{7.3.11}{4.3.2}\)
\(A=\frac{7.11}{4.2}\)
\(A=\frac{77}{8}\)
A=7/4.(3333/1212+3333/2020+3333/3030+3333/4242)
A=7/4.(33/12+33/20+33/30+33/42)
A=7/4.33.(1/3*4+1/4*5+1/5*6+1/6*7)
A=231/4.(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)
A=231/4.(1/3-1/7)
A=231/4.4/21
A=11. Vay A=11
Nho k cho minh voi nhe
A= 7/4-(33/12x101+33/20x101+33/30x101+33/42x101)
=7/4-[101x(33/12+33/20+33/30+33/42)]
=7/4-44/7
=-127/28
\(A=\frac{7}{4}\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
\(A=\frac{7}{4}\left(\frac{385}{140}+\frac{231}{140}+\frac{154}{140}+\frac{110}{140}\right)\)
\(A=\frac{7}{4}.\frac{44}{7}\)
\(A=\frac{44}{4}=11\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
\(\frac{7}{4}.\left(\frac{101.33}{101.12}+\frac{101.33}{101.20}+\frac{101.33}{101.30}+\frac{101.33}{101.42}\right)\)
\(=\frac{7.33}{4}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\\ =\frac{7.33}{4}\left(\frac{35+21+14+1}{420}\right)\)
\(=\frac{7.3.11}{4}.\frac{71}{420}=\frac{7.3.11.71}{4.4.5.3.7}=\frac{781}{100}\)
mk lm chak vớ vẩn rồi
Vì A > 1; B < 1 nên A > B.
bao quynh Cao bạn ơi hình như bn làm sai đề ạ 7/4 mà sao lại 4/7 ạ