Question

[Bùi Công Tiến Anh]

Bài 1: Cho Sn= ( 2 - \(\sqrt{3}\))ⁿ + ( 2 + \(\sqrt{3}\))ⁿ     ( n thuộc Z⁺ )

    1) CM: S3n + 3Sn = Sn³

       2)  Tính S₃, S₄​

Bài 2: Tìm Min

F= x⁴ + 6x³ + 13x² +12x + 12

Answer 1

[Incursion_03]

\(B1,1,S_{3n}+3S_n=\left(2-\sqrt{3}\right)^{3n}+\left(2+\sqrt{3}\right)^{3n}+3\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]\)

         \(=\left[\left(2-\sqrt{3}\right)^n\right]^3+\left[\left(2+\sqrt{3}\right)^n\right]^3\)

                         \(+3\left[\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\right]^n\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]\)

Ta có hằng đẳng thức \(a^3+b^3+3ab\left(a+b\right)=\left(a+b\right)^3\)

Ở đây với \(a=\left(2-\sqrt{3}\right)^n\)và \(b=\left(2+\sqrt{3}\right)^n\)

Nên \(S_{3n}+3S_n=\left[\left(2-\sqrt{3}\right)^n+\left(2+\sqrt{3}\right)^n\right]^3=S_n^3\)

\(2,S_3=\left(2-\sqrt{3}\right)^3+\left(2+\sqrt{3}\right)^3\)

         \(=\left(2-\sqrt{3}+2+\sqrt{3}\right)\left(2-\sqrt{3}-\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)+2+\sqrt{3}\right)\)

        \(=4\left[4-\left(4-3\right)\right]\)

         \(=12\)

Ta có \(S_4=\left(2-\sqrt{3}\right)^4+\left(2+\sqrt{3}\right)^4\)

              \(=\left[\left(2-\sqrt{3}\right)^2\right]^2+\left[\left(2+\sqrt{3}\right)^2\right]^2\)

              \(=\left(7-4\sqrt{3}\right)^2+\left(7+4\sqrt{3}\right)^2\)

              \(=97-56\sqrt{3}+97+56\sqrt{3}\)

              \(=194\)

Answer 2

[Incursion_03]

\(B2,F=x^4+6x^3+13x^2+12x+12\)(Bài này cẩn thận dấu "=")

            \(=\left(x^4+6x^3+9x^2\right)+4x^2+12x+12\)

            \(=\left(x^2+3x\right)^2+4\left(x^2+3x\right)+4+8\)

             \(=\left(x^2+3x+2\right)^2+8\ge8\)

Dấu "=" tại \(x^2+3x+2=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)