\(\frac{\frac{2}{7}+\frac{2}{9}-\frac{2}{121}}{\frac{4}{7}+\frac{4}{9}-\frac{4}{121}}=\frac{2.\hept{ }\frac{1}{7}+\frac{1}{9}-\frac{1}{121}}{4.\hept{ }\frac{1}{7}+\frac{1}{9}-\frac{1}{121}}=\frac{2}{4}=\frac{1}{2}\)

bạn nhớ thay ngoặc nhọn thành ngoặc tròn nhé và nhớ đóng mở ngoặc

dễ thế mà ko bk làm hả em

dễ thế mà ko biết

\(-\frac{4}{7}\)Tối giản gòi sau rút gọn dc

=\(\frac{-4}{7}\)

kb nha 

\(=\frac{-4}{7}\)

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

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

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

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

\(B=\frac{1}{\sqrt{5}+\sqrt{7}}-\frac{1}{\sqrt{5}-\sqrt{7}}=\frac{\sqrt{5}-\sqrt{7}-\sqrt{5}-\sqrt{7}}{5-7}=\frac{-2\sqrt{7}}{-2}=\sqrt{7}\)

\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}=\sqrt{\left(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)^2}\)

\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}+2\sqrt{\frac{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}}+\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)

\(C=\sqrt{\frac{\left(4+\sqrt{7}\right)^2}{16-7}+\frac{\left(4-\sqrt{7}\right)^2}{16-7}+2}\)

\(C=\sqrt{\frac{\left(4+\sqrt{7}+4-\sqrt{7}\right)^2-2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{16-7}+2}\)

\(C=\sqrt{\frac{16^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{238}{9}+2}=\sqrt{\frac{256}{9}}=\frac{16}{3}\)

Chúc bạn học tốt ~ 

thanks ban 

đoạn cuối sửa lại nhé -,- tính ngu 

\(C=\sqrt{\frac{8^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{46}{9}+2}=\sqrt{\frac{64}{9}}=\frac{8}{3}\)

Chúc bạn học tốt ~ 

A=\(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\) 

A=\(\frac{7}{4}.\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\) 

A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)

A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3-4}+\frac{1}{4-5}+\frac{1}{5-6}+\frac{1}{6-7}\right)\right]\)

A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\) 

A=\(\frac{7}{4}.\frac{44}{7}\)

A=11

Like cho mình nha bài này viết mỏi tay lắm