\(\frac{\left(\frac{3}{15}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{5\frac{1}{3}+\frac{2}{5}}\)

\(=\frac{\left(\frac{12}{60}+\frac{15}{60}+\frac{21}{60}\right)\times\frac{17}{49}}{\frac{16}{3}\times\frac{2}{5}}\)

\(=\frac{\frac{48}{60}\times\frac{17}{49}}{\frac{80}{15}+\frac{6}{15}}\)

\(=\frac{\frac{816}{2940}}{\frac{86}{15}}\)

\(=\frac{816}{2940}:\frac{86}{15}\)

\(=\frac{816}{2940}\times\frac{15}{86}\)

\(=\frac{68}{245}\times\frac{15}{86}\)

\(=\frac{102}{2107}\)

a)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A =  - 1\end{array}\)

b)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)

\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)

 \(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+....+\frac{50^2}{49.51}\)

\(=\frac{2^2-1}{1.3}+\frac{3^2-1}{2.4}+....+\frac{50^2-1}{49.51}+\frac{1}{1.3}+\frac{1}{2.4}+....+\frac{1}{49.51}\)

\(=\frac{1}{2}.\left(1+1+...+1\right)+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\)

Tự làm tiếp :)) 

tớ nhầm đoạn này tí :((

\(=\left(1+1+....+1\right)+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)(49 chữ số 1)

\(=49+\frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\right)\right]\)

\(=49+\left(\frac{3}{2}-\frac{1}{50}-\frac{1}{51}\right):2\)Tự tính 

sao bạn Khuyển Dạ Xoa làm 2 bài vậy?

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)

\(B=\frac{1}{4}\)

a) \(25^{\dfrac{1}{2}}=5\)

b) \(\left(\dfrac{36}{49}\right)^{-\dfrac{1}{2}}=\dfrac{7}{6}\)

c) \(100^{1,5}=1000\)

P= \(\frac{1}{3}\)+\(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+......+\frac{1}{1275}\)

Ta nhân tất cả phân số với 2/2 và không rút gọn

P = \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}\)\(+\)\(......+\frac{2}{2550}\)

Ta có công thức:

\(\frac{a}{b.c}=\frac{a}{c-b}.\left[\frac{1}{b}-\frac{1}{c}\right]\)

=> P = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{50.51}\)

P = \(2.\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{50}-\frac{1}{51}\right]\)

\(P=2.\left[\frac{1}{2}-\frac{1}{51}\right]\)

\(P=2.\frac{49}{102}\)\(=\frac{49}{51}\)

Đó là cách làm của tớ, có gì không hiểu rạng sáng ngày 18 tháng 3 hỏi nhé!

mình cũng chịu

\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1}{4}\)