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[OSP 2015 Esai No 2] Sistem persamaan siklis

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Tentukan semua tripel bilangan real $(x,y,z)$ yang memenuhi sistem persamaan


$$ (x+1)^2=x+y+2$$


$$(y+1)^2=y+z+2$$


$$(z+1)^2=z+x+2 $$


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jika \(x=-1\) maka\(y=-1\) dan \(z=-1\).

jika \(x\neq -1\)

perhatikan bahwa sistem jadi

\(x^2+x=y+1\)

dst.

kalikan ketiganya, didapat \(xyz=1\).

jumlahkan ketiganya didapat \(x^2+y^2+z^2 =3\). Karena ruas kiri positive maka 

\[ x^2+y^2+z^2 \geq 3xyz\]

Hal ini benar apalagi jika ruas kanan negatif.

akibatnya \(3=3\) dan kesamaan tercapai.

Jadi \(x^2=y^2=z^2\) dan substitusi balik didapat \( x=y=z=\pm 1\).

 

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\begin{align*}  (x+1)^2 &=x+y+2 \\ x^2 + 2x +1 &= x + y +2 \\ \therefore x^2 + x &= y+1 ...(1 \\ (y+1)^2 &=y+z+2 \\ y^2 + 2y +1 &= y+z+2  \\  \therefore  y^2 + y &=z +1 ...(2 \\ (z+1)^2 &=z + x + 2 \\ z^2 + 2z + 1 &= z +x +2 \\  \therefore z^2 + z &= x +1 ...(3 \end{align*}

Kalikan ketiga persamaan.

\begin{align*} (x^2 + x)(y^2 + y)(z^2 + z) &= (y+1)(z+1)(x+1) \\ x(x+1) \cdot y(y+1) \cdot z(z+1) &= (x+1)(y+1)(z+1) \\ \therefore xyz &= 1 \end{align*}

Jumlahkan ketiga persamaan.

\begin{align*} (x^2 +x) + (y^2 + y) + (z^2 + z) &= (y +1) + (z + 1) + (x + 1) \\ x^2 + y^2 + z^2 + x + y + z&= x + y + z + 3 \\ \therefore x^2 + y^2 + z^2 &=3 \end{align*}

Perhatikan bahwa $x^2 + y^2 + z^2 \ge 0$. Sehingga menyebabkan

\begin{align*} x^2 + y^2 + z^2 &\ge 3xyz \\ \Rightarrow x^2 + y^2 + z^2 &\ge 3 \end{align*}

Kesamaan terjadi jika $x=y=z$, maka diperoleh nilai $x=y=z=\pm 1$.

 

Edited by Wildan Bagus W

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