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Kudo Shinichi · Accepted Answer

$b,\sqrt{36}.\sqrt{\dfrac{25}{26}}+\dfrac{1}{4}\
=\sqrt{6^2}.\sqrt{\left(\dfrac{5}{4}\right)^2}+\dfrac{1}{4}\=6.\dfrac{5}{4}+\dfrac{1}{4}=\dfrac{30}{4}+\dfrac{1}{4}=\dfrac{31}{4}$$c,\sqrt{\dfrac{4}{81}}:\sqrt{\dfrac{25}{81}}-1\dfrac{2}{5}\ =\sqrt{\left(\dfrac{2}{9}\right)^2}:\sqrt{\left(\dfrac{5}{9}\right)^2}-\dfrac{7}{5}\ =\dfrac{2}{9}:\dfrac{5}{9}-\dfrac{7}{5}\ =\dfrac{2}{9}.\dfrac{9}{5}-\dfrac{7}{5}=\dfrac{2}{5}-\dfrac{7}{5}\ =-1$$d,
0,1.\sqrt{225}.\sqrt{\dfrac{1}{4}}\
=\dfrac{1}{10}.\sqrt{15^2}.\sqrt{\left(\dfrac{1}{2}\right)^2}\
=\dfrac{1}{10}.15.\dfrac{1}{2}=\dfrac{3}{5}$$e,
\dfrac{3^{25}}{9^3.3^{16}}\
=\dfrac{3^{25}}{\left(3^2\right)^3.3^{16}}\
=\dfrac{3^{25}}{3^6.3^{16}}\
=\dfrac{3^{25}}{3^{22}}\
=3^3=27$Câu trả lời: $b,\sqrt{36}.\sqrt{\dfrac{25}{26}}+\dfrac{1}{4}\
=\sqrt{6^2}.\sqrt{\left(\dfrac{5}{4}\right)^2}+\dfrac{1}{4}\=6.\dfrac{5}{4}+\dfrac{1}{4}=\dfrac{30}{4}+\dfrac{1}{4}=\dfrac{31}{4}$$c,\sqrt{\dfrac{4}{81}}:\sqrt{\dfrac{25}{81}}-1\dfrac{2}{5}\ =\sqrt{\left(\dfrac{2}{9}\right)^2}:\sqrt{\left(\dfrac{5}{9}\right)^2}-\dfrac{7}{5}\ =\dfrac{2}{9}:\dfrac{5}{9}-\dfrac{7}{5}\ =\dfrac{2}{9}.\dfrac{9}{5}-\dfrac{7}{5}=\dfrac{2}{5}-\dfrac{7}{5}\ =-1$$d,
0,1.\sqrt{225}.\sqrt{\dfrac{1}{4}}\
=\dfrac{1}{10}.\sqrt{15^2}.\sqrt{\left(\dfrac{1}{2}\right)^2}\
=\dfrac{1}{10}.15.\dfrac{1}{2}=\dfrac{3}{5}$$e,
\dfrac{3^{25}}{9^3.3^{16}}\
=\dfrac{3^{25}}{\left(3^2\right)^3.3^{16}}\
=\dfrac{3^{25}}{3^6.3^{16}}\
=\dfrac{3^{25}}{3^{22}}\
=3^3=27$

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