Номера 24,25,26,27,28 огромное спасибо даю 25 баллов!

[latex]23)\quad a- \frac{1}{a} = \frac{2}{3} \; \; \to (a- \frac{1}{a} )^2= \frac{4}{9} \; \; \to \\\\a^2-2\cdot a\cdot \frac{1}{a} + \frac{1}{a^2} = \frac{4}{9} \\\\\underbrace {a^2+\frac{1}{a^2}}-2=\frac{4}{9}\; \; \; \to \; \; \; a^2+\frac{1}{a^2}=2+\frac{4}{9}\\\\a^2+\frac{1}{a^2}=\frac{22}{9}=2\frac{4}{9}\\\\\frac{a^4+1}{a^2} = \frac{a^4}{a^2} + \frac{1}{a^2} =a^2+\frac{1}{a^2}=2\frac{4}{9}\\\\24)\quad a+\frac{1}{a}=3\; \; \; \to \; \; \; (a+\frac{1}{a})^3=27\; \; \to \\\\a^3+3a^2\cdot \frac{1}{a}+3a\cdot \frac{1}{a^2}+\frac{1}{a^3}=27[/latex] [latex]a^3+3a+\frac{3}{a}+\frac{1}{a^3}=27[/latex]   [latex]a^3+\frac{1}{a^3}=27-3a-3\cdot \frac{1}{a}=27-3(a+\frac{1}{a})=27-3\cdot 3=27-9=18[/latex]   [latex]\frac{a^6+1}{a^3}=\frac{a^6}{a^3}+\frac{1}{a^3}=a^3+\frac{1}{a^3}=18\\\\25)\quad a-\frac{1}{a}=\sqrt7\; \; \; \to \; \; \; (a-\frac{1}{a})^2=7\; \; \; \to \\\\a^2-2a\cdot \frac{1}{a}+\frac{1}{a^2}=a^2-2+\frac{1}{a^2}=7\; \; \to \\\\a^2+\frac{1}{a^2}=7+2=9\; \; \; \to \; \; \; (a^2+\frac{1}{a^2})^2=9^2\\\\(a^2+\frac{1}{a^2})^2=a^4+2+\frac{1}{a^4}=81\; \; \; \to \; \; \; a^4+\frac{1}{a^4}=81-2=79[/latex]