Bài 1: Tính:
a) \(\sqrt {{{( – 3)}^4}} – \sqrt {{{\left( { – 7} \right)}^2}} + \sqrt { – {{\left( { – 4} \right)}^3}} \)
b) \(\left| { – {3 \over 7}} \right|:{\left( { – 3} \right)^2} – \sqrt {{4 \over {49}}} \)
Bài 2: So sánh: \(A = {222^{555}}\) và \(B = {555^{222}}\).
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Bài 1: a) \(\sqrt {{{( – 3)}^4}} – \sqrt {{{\left( { – 7} \right)}^2}} + \sqrt { – {{\left( { – 4} \right)}^3}} \)
\(\;= \sqrt {81} – \sqrt {49} + \sqrt {64} \)
\(\;= 9 – 7 + 8 = 10\)
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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 \(A > B.\)
