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r091995 @r091995

October 2021 2 22 Report

Упростите выражение

[tex]\frac{5}{4+\sqrt{11}}+\frac{8}{\sqrt{19}-\sqrt{11}}-\frac{10}{\sqrt{19}+3}[/tex]

Вычислите :

[tex]9^{\frac{3}{2}}+27^{\frac{2}{3}}-(\frac{1}{16})^{-\frac{3}{4}}[/tex]

Artem112

Verified answer

$\frac{5}{4+\sqrt{11}}+\frac{8}{\sqrt{19}-\sqrt{11}}-\frac{10}{\sqrt{19}+3}=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{8(\sqrt{19}+\sqrt{11})}{(\sqrt{19}-\sqrt{11})(\sqrt{19}+\sqrt{11})}-\frac{10(\sqrt{19}-3)}{(\sqrt{19}+3)(\sqrt{19}-3)}=\frac{5(4-\sqrt{11})}{5}+\frac{8(\sqrt{19}+\sqrt{11})}{8}-\frac{10(\sqrt{19}-3)}{10}=4-\sqrt{11}+\sqrt{19}+\sqrt{11}-\sqrt{19}+3=4+3=7$

$9^{\frac{3}{2}}+27^{\frac{2}{3}}-(\frac{1}{16})^{-\frac{3}{4}}=9^{\frac{3}{2}}+27^{\frac{2}{3}}-16^{\frac{3}{4}}=\sqrt[2]{9^3} +\sqrt[3]{27^2} -\sqrt[4]{16^3} =\sqrt[2]{9^3} +\sqrt[3]{27^2} -\sqrt[4]{16^3} =3^3+3^2-2^3=27+9-8=28$

2 votes Thanks 1

kiskam

вложил решение ниже.....

1 votes Thanks 1

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