Cac ban giup mik voi, mik k cho! ( 10 k cho 3 nguoi dau tien tra loi cau hoi cua mik)

tự tính  nhé dễ lắm bạn 

đềlà gì vậy

a) \(\frac{-6}{21}.\frac{3}{2}=-\frac{3}{7}\)          b) \(\left(-3\right).\left(\frac{-7}{12}\right)=\frac{21}{12}=\frac{7}{4}\)

c) \(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}=\frac{11}{12}.\frac{16}{33}.\frac{3}{5}=\frac{4}{15}\)

d) \(\sqrt{\left(-7\right)^2}+\sqrt{\frac{2}{16}}=7+\sqrt{\frac{1}{8}}\)

c) \(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0=\frac{1}{2}.10-\frac{1}{4}+1=5\frac{3}{4}\)

Ta có:

\(A=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{5}{12}-\frac{1}{16}}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{17}{48}}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{51}}{12}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{17}}{12}\)

\(A=\frac{4\sqrt{3}+2\sqrt{2}+\sqrt{17}}{12}\)

Ta có: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{6}}=\sqrt{\frac{5-2\sqrt{6}}{12}}\)

Vì \(5-2\sqrt{6}=3-2\sqrt{3}.\sqrt{2}+2=\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\)\(\Rightarrow5-2\sqrt{6}=\left(\sqrt{3}-\sqrt{2}\right)^2\)

Như vậy: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)

Lại có: \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}.\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)

Rút gọn ta được \(A=\frac{\sqrt{3}}{2}\)

Cây a, bạn nhân cả 2 vế với 3

Lấy vế nhân với 3 trừ đi ban đầu tất cả chia 2

b) Tính như bình thường

Câu c hình như sai đề

a) \(1-\frac{1}{3^{101}}\)

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)

\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)

\(=\frac{1454}{323}+\frac{35}{43}+6\)

\(=5,...+6\)

\(=11,...\)

\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)

\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)

\(=\sqrt{3}-2\) 

\(VayA=\sqrt{3}-2\)

A = \(\frac{\frac{\frac{5+3.3-1.12}{12}}{3.6-5+2.2}+}{6}+\frac{16\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}=\frac{\frac{\frac{5+9-12}{12}}{18-5+4}}{6}+\frac{16}{17}=\frac{2}{12}.\frac{6}{17}+\frac{16}{17}=\frac{1}{17}.\frac{6}{17}=1\)

A=\(\frac{\frac{5}{12}+\frac{9}{12}-\frac{12}{12}}{\frac{18}{6}-\frac{5}{6}+\frac{4}{6}}+\frac{16.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}\)

=\(\frac{\frac{1}{6}}{\frac{17}{6}}+\frac{16}{17}\)

=\(\frac{1}{17}+\frac{16}{17}\)

=1