Bài 1:

\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)

\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)

\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)

\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)

Bài 2: 

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)

B = 3/1.4 + 5/4.9 + 7/9.16 + 15/16.31 + 19/31.50

Ta co cong thuc : 1/n.(n-1)=1/n-1/n-1

=> B = 1/1 - 1/4 +1/4 - 1/9 + 1/9 - 1/16 + 1/16 - 1/31 + 1/31 - 1/50

        = 1/1 - 1/50

        = 49/50

\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+\frac{15}{16.31}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}\)

\(=1-\frac{1}{31}\)

\(=\frac{30}{31}\)

Dựa vào công thức được chứng minh:

(Em có thể chứng minh lại)

Ta có:

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}\)

\(=1-\frac{1}{31}\)

\(=\frac{30}{31}\)

Chúc em học tốt^^

bạn hãy thử nghĩ ra cách tính nhanh nếu có thể :)) mới vô bạn phải nghĩ đến việc 3=4-1 để rút được cái tử số và các số 5,7,15 cũng tách như vậy dựa theo mẫu

cuối cùng ta được \(\frac{4}{1.4}-\frac{1}{1.4}+\frac{9}{4.9}-\frac{4}{4.9}+...\)... Mấy cái sau bạn tự khai triển nhá

Rút lại ta được 1-\(\frac{1}{31}\)là bằng 30 phần 31

\(\text{Ta có :}\)

\(\frac{3}{1.4}=1-\frac{1}{4}\)

\(\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\)

\(\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\)

\(......\)

\(\frac{19}{81.100}=\frac{1}{81}-\frac{1}{100}\)

\(\text{Cộng vế với vế ta có:}\)

\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

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

Ko biết có đc k ko ta!?

Nguyễn Xuân Anh cảm ơn cậu nha ^^

Cảm ơn mà toàn k sai hoài là sao!! ahuhu

\(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)

\(C=\frac{7^2}{7}.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)

\(C=7.\left(\frac{1}{2}-\frac{1}{72}\right)\)

\(C=7.\frac{35}{72}=\frac{245}{72}\)

Ta có : \(C=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+.....+\frac{7^2}{65.72}\)

\(\Rightarrow C=7\left(\frac{7}{2.9}+\frac{7}{9.16}+\frac{7}{16.23}+.....+\frac{7}{65.72}\right)\)

\(\Rightarrow C=7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.....+\frac{1}{65}-\frac{1}{72}\right)\)

\(\Rightarrow C=7\left(\frac{1}{2}-\frac{1}{72}\right)\)

\(\Rightarrow C=7.\frac{35}{72}=\frac{245}{72}\)

\(C=7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-.....+\frac{1}{65}-\frac{1}{72}\right)..\)

\(C=7\left(\frac{1}{2}-\frac{1}{72}\right)..\)

\(C=?\)

a)

\(A=\left(\frac{1}{9}-\frac{1}{10}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-....-\left(1-\frac{1}{2}\right)=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-....-1+\frac{1}{2}\)

\(A=-\left(\frac{1}{10}+1\right)=-\frac{11}{10}\)

a)\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\\ \Rightarrow A=-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\\ \Rightarrow A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)Đặt \(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\)

\(\Rightarrow B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow B=1-\frac{1}{10}=\frac{9}{10}\)

Ta có : \(A=-B\)

\(\Rightarrow A=-\frac{9}{10}\)

a) A=\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

A=\(-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)

A=\(-\left(\frac{1}{9.10}+\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)

A=-\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{9}-\frac{1}{10}\right)\)

A=-(\(1-\frac{1}{10}\))

A=-\(\frac{9}{10}\)

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=\dfrac{2^3}{1}=2^3=8\)

_____

\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)

`=`\(\dfrac{\left(2^2\right)^8\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}\)

`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)

`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}\)

`=`\(\dfrac{3^2}{2}\) `=`\(\dfrac{9}{2}\)

______

\(\dfrac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)

`=`\(\dfrac{\left(3^3\right)^4\cdot2^3-3^{10}\cdot\left(2^2\right)^3}{2^4\cdot3^4\cdot\left(3^2\right)^3}\)

`=`\(\dfrac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^4\cdot3^4\cdot3^6}\)

`=`\(\dfrac{3^{10}\cdot\left(3^2\cdot2^3-2^6\right)}{3^{10}\cdot2^4}\)

`=`\(\dfrac{72-2^6}{2^4}=\dfrac{8}{16}=\dfrac{1}{2}\)

\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=2^3=8\)

\(\dfrac{4^8.9^4}{6^6.8^3}=\dfrac{2^{16}.3^8}{2^6.3^6.2^9}=2.3^2=18\)

\(\dfrac{27^4.2^3-3^{10}.4^3}{6^4.9^3}=\dfrac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6}=\dfrac{2^3.3^{10}.\left(3^2-2^3\right)}{2^4.3^{10}}=\dfrac{9-8}{2}=\dfrac{1}{2}\)