\(x=\frac{903}{391}\)

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903/391 mình nghĩ vậy

x = 903/391

đề bài là:(3/20+1/2-1/5)-12/49 phần 10/3 dổi từ3mot phan ba +2/9 đúng ko

\(\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{5}\right)-\frac{12}{49}=\left(\frac{3}{20}+\frac{10}{20}-\frac{4}{20}\right)-\frac{12}{49}=\frac{9}{20}-\frac{12}{49}=\frac{201}{980}\)

\(3\frac{1}{3}+\frac{2}{9}=\frac{10}{3}+\frac{2}{9}=\frac{30}{9}+\frac{2}{9}=\frac{32}{9}\)

 tick đúng cho mk nha tran Thi Thanh Hong 01

\(\frac{\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{5}\right)-\frac{12}{49}}{\frac{10}{3}+\frac{2}{9}}=\frac{\left(\frac{3}{20}+\frac{10}{20}-\frac{4}{20}\right)-\frac{12}{49}}{\frac{30}{9}+\frac{2}{9}}=\frac{\frac{9}{20}-\frac{12}{49}}{\frac{32}{9}}=\frac{\frac{201}{980}}{\frac{32}{9}}=....\)

đáp số ra hơi lớn

\(\frac{\left(\frac{3}{15}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{5\frac{1}{3}+\frac{2}{5}}=\frac{\left(\frac{1}{5}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{\frac{16}{3}+\frac{2}{5}}\)

=\(\frac{\left(\frac{4}{20}+\frac{5}{20}+\frac{7}{20}\right)\times\frac{17}{49}}{\frac{80}{15}+\frac{6}{15}}=\frac{\frac{16}{20}\times\frac{17}{49}}{\frac{86}{15}}=\frac{\frac{4}{5}\times\frac{17}{49}}{\frac{86}{15}}\)

=\(\frac{68}{245}\times\frac{15}{86}=\frac{102}{2107}\)

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

Mình đang cần gắp 

bạn còn

a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)

b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0

a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)

\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)

\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(=\frac{2}{3}+\frac{1}{11}\)

\(=\frac{25}{33}\)

b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)

Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.

Ta có \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)

 = \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)

= \(\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

=     \(\frac{2}{3}+\frac{1}{11}\)

=       \(\frac{25}{33}\)

Ta có nếu theo quy luật như trên thì sẽ có 1 thừa số là\(\frac{1}{49}-\frac{1}{7^2}\)

Mà chúng bằng 0 nên tích trên bằng 0