Question

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`d)``(3-a)^2``-``\sqrt{0,2}``.``\sqrt{180.a^2}`

Answer 1

\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180.a^2}\\ =9-6a+a^2-\sqrt{0,2.180}.\sqrt{a^2}\\ =9-6a+a^2-\sqrt{36}.\left|a\right|\\ TH_1:a>0\\ 9-6a+a^2-6a\\ =9-12a+a^2\\ TH_2:a< 0\\ 9-6a+a^2-6.\left(-a\right)\\ =9-6a+a^2+6a\\ =9+a^2\)

Answer 2

\(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180\cdot a^2}\)

\(=3^2-6a+a^2-\sqrt{0,2}\cdot a\sqrt{180}\)

\(=3^2-6a+a^2-\dfrac{\sqrt{5}}{5}\cdot a\cdot6\sqrt{5}\)

\(=9-6a+a^2-6a\)

\(=a^2-12a+9\)

Tự giải nghiệm.