HOTS Zone : Persamaan Aljabar

\begin{aligned} \dfrac{x^2+y^2}{x^2-y^2}+\dfrac{x^2-y^2}{x^2+y^2}&=3\\ \dfrac{\left(x^2+y^2\right)^2+\left(x^2-y^2\right)^2}{\left(x^2+y^2\right)\left(x^2-y^2\right)}&=3\\ \dfrac{x^4+2x^2y^2+y^4+x^4-2x^2y^2+y^4}{x^4-y^4}&=3\\ \dfrac{2x^4+2y^4}{x^4-y^4}&=3\\ 2x^4+2y^4&=3x^4-3y^4\\ 5y^4&=x^4\\ x^8&=25y^8 \end{aligned} \begin{aligned} \dfrac{x^{8}+y^{8}}{x^{8}-y^{8}}+\dfrac{x^{8}-y^{8}}{x^{8}+y^{8}}&=\dfrac{25y^8+y^{8}}{25y^8-y^{8}}+\dfrac{25y^8-y^{8}}{25y^8+y^{8}}\\ &=\dfrac{26y^8}{24y^8}+\dfrac{24y^8}{26y^{8}}\\ &=\dfrac{13}{12}+\dfrac{12}{13}\\ &=\dfrac{169+144}{156}\\ &=\dfrac{313}{156} \end{aligned} a = 313, b = 156

a + b = 313 + 156 = 469469

\begin{aligned} x^2+\dfrac{x^2}{(x+1)^2}&=3\\ x^2-2\cdot x\cdot\dfrac{x}{x+1}+\left(\dfrac{x}{x+1}\right)^2+2\cdot x\cdot\dfrac{x}{x+1}&=3\\ \left(x-\dfrac{x}{x+1}\right)+2\cdot\dfrac{x^2}{x+1}&=3\\ \left(\dfrac{x(x+1)-x}{x+1}\right)+2\cdot\dfrac{x^2}{x+1}-3&=0\\ \left(\dfrac{x^2+x-x}{x+1}\right)+2\cdot\dfrac{x^2}{x+1}-3&=0\\ \left(\dfrac{x^2}{x+1}\right)+2\cdot\dfrac{x^2}{x+1}-3&=0\\ \end{aligned} Misal $\dfrac{x^2}{x+1}=p$
\begin{aligned} p^2+2p-3&=0\\ (p+3)(p-1)&=0 \end{aligned}

• p + 3 = 0
  \(\begin{aligned} p&=-3\\ \dfrac{x^2}{x+1}&=-3\\[4pt] x^2&=-3x-3\\ x^2+3x+3&=0 \end{aligned}\)
  
  \(\begin{aligned} D&=3^2-4(1)(3)\\ &=9-12\\ &=-3 \end{aligned}\)
  Karena D < 0 sehingga persamaan di atas tidak ada solusi real.
• p − 1 = 0
  \(\begin{aligned} p&=1\\ \dfrac{x^2}{x+1}&=1\\[4pt] x^2&=x+1\\ x^2-x-1&=0\\ x&=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}\\[4pt] &=\dfrac{1\pm\sqrt{1+4}}2\\ &=\boxed{\boxed{\dfrac{1\pm\sqrt5}2}} \end{aligned}\)

x > 0

\(\begin{aligned} \dfrac{x^2}{10000}&=\dfrac{10^5}{x^{2\left({^{10}\negmedspace\log x}\right)-8}}\\[4pt] \dfrac{x^2}{10000}&=\dfrac{10^5}{x^{2\log x-8}}\\[4pt] x^2\cdot x^{2\log x-8}&=10^5\cdot10^5\\ x^{2+2\log x-8}&=10^{5+5}\\ x^{2\log x-6}&=10^{10}\\ x^{\frac{2\log x-6}2}&=10^{\frac{10}2}\\ x^{\log x-3}&=10^5\\ \log\left(x^{\log x-3}\right)&=\log10^5\\ \left(\log x-3\right)\log x&=5 \end{aligned}\)
Misal log x = p
\(\begin{aligned} (p-3)p&=5\\ p^2-3p-5&=0 \end{aligned}\)

\(\begin{aligned} p_1+p_2&=\dfrac{-(-3)}1\\[4pt] \log x_1+\log x_2&=3\\ \log\left(x_1\cdot x_2\right)&=3\\ x_1\cdot x_2&=\boxed{\boxed{10^3}} \end{aligned}\)

\(\begin{aligned} ab&\leq\dfrac{(a+b)^2}4\\[4pt] &\leq\dfrac{8^2}4\\[4pt] &\leq16 \end{aligned}\)

\(\begin{aligned} c^2+16&\geq16\\ ab&\geq16\\ \end{aligned}\)
Didapat ab = 16 dan c = 0

a + b + c = 8 + 0 = 8