Ta có : \(C=\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+...+\frac{1}{2652}=\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+..+\frac{1}{51.52}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{51}-\frac{1}{52}\)

\(=\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{51}+\frac{1}{52}\right)-2\left(\frac{1}{8}+\frac{1}{10}+...+\frac{1}{52}\right)\)

\(=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{51}+\frac{1}{52}-\left(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{26}\right)\)\(=\frac{1}{27}+\frac{1}{28}+...+\frac{1}{52}\)

Khi đó ta không thể chứng minh C < 1/4 vì sở dĩ \(\frac{1}{27}+\frac{1}{28}+...+\frac{1}{34}>\frac{1}{4}\)(bạn thử lấy máy tính tính xem) 

\(\left(1-\frac{2}{42}\right)\left(1-\frac{2}{56}\right)...\left(1-\frac{2}{2652}\right)\)

= \(\frac{40}{42}.\frac{54}{56}.\frac{70}{72}...\frac{2650}{2652}\)

= \(\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}...\frac{50.53}{51.52}\)

= \(\frac{5.6.7...50}{6.7.8...51}.\frac{8.9.10...53}{7.8.9...52}\)

= \(\frac{5}{51}.\frac{53}{7}=\frac{265}{357}\)

b, \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

.............................................

Cộng với vế 99 của BĐT trên, ta được:

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>99.\frac{1}{10}=\frac{99}{10}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}=\frac{1}{10}=\frac{100}{10}=10\)

Wrecking Ball đã làm đúng

to ra kết quả giống cậu : Wrecking Ball

là đáp án đúng

tk nha ( chúc bn học gioi )

k cho đi chế ơi

A = 1/20 + 1/72 = 23/360

B = 1/2 + 1/6 + 1/30 = 7/10

C = 1/42 + 1/56 + 1/12 = 1/8

Trung bình cộng của A, B và C là: (23/360 + 7/10 + 1/8) : 3 = 8/27

Đáp số: 8/27

a)Đặt \(A=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)

\(A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)

\(A=\dfrac{1}{3}-\dfrac{1}{12}\)

\(A=\dfrac{1}{4}\)

b)Đặt \(B=\dfrac{1}{501}+\dfrac{1}{502}+...+\dfrac{1}{1000}\)(có 500 số hạng)

\(B< \dfrac{1}{500}+\dfrac{1}{500}+...+\dfrac{1}{500}\)(có 500 số hạng)

\(B< 500\cdot\dfrac{1}{500}=1\)

\(\Rightarrow B< 1\left(đpcm\right)\)

A=1/2+1/6+....+1/56+1/72

A=1/1.2+1/2.3+...+1/7.8+1/8.9

A=1/1-1/2+1/2-1/3+...+1/7-1/8+1/8-1/9

A=1/1-1/9=9/9-1/9=8/9

Thanks ban nha

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..\frac{1}{8.9}\)

A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{8}-\frac{1}{9}\)

A=1-\(\frac{1}{9}\)=\(\frac{8}{9}\)

\(M=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

=> \(M=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)

=> \(M=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

=> \(M=\frac{1}{5}-\frac{1}{12}=\frac{12}{60}-\frac{5}{60}=\frac{7}{60}>\frac{7}{70}=\frac{1}{7}\)

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