Bài 1:
a) \(\sqrt {{{( - 3)}^4}} - \sqrt {{{\left( { - 7} \right)}^2}} + \sqrt { - {{\left( { - 4} \right)}^3}} \)
\(\;= \sqrt {81} - \sqrt {49} + \sqrt {64} \)
\(\;= 9 - 7 + 8 = 10\)
b) \(\left| { - {3 \over 7}} \right|:{\left( { - 3} \right)^2} - \sqrt {{4 \over {49}}} \)
\(\;= {3 \over 7}:9 - {2 \over 7} = {3 \over {63}} - {2 \over 7} \)
\(\;= {{3 - 18} \over {63}} = {{ - 15} \over {63}} = {{ - 5} \over {21}}.\)
Bài 2: Ta có: \(A = {222^{555}} = {\left( {2.111} \right)^{555}} = {2^{555}}{.111^{555}} \)\(\;= {\left( {{2^5}} \right)^{111}}{.111^{555}} = {32^{111}}{.111^{555}}\)
\(B = {555^{222}} = {\left( {5.111} \right)^{222}} = {5^{222}}{.111^{222}} \)\(\;= {\left( {{5^2}} \right)^{111}}{.111^{222}} = {25^{111}}{.111^{222}}\)
Vì \({32^{111}} > {25^{111}}\) và \({111^{555}} > {111^{222}}\)
Nên \({32^{111}}{.111^{555}} > {25^{111}}{.111^{222}}\) hay \(A > B.\)
