HOTS Zone : Aljabar [6]

\(\begin{aligned} x^2+\dfrac{x^2}{(x+1)^2}&=3\\ (x+1-1)^2+\dfrac{(x+1-1)^2}{(x+1)^2}&=3\\ (x+1)^2-2(x+1)+1+\dfrac{(x+1)^2-2(x+1)+1}{(x+1)^2}&=3\\ (x+1)^2-2(x+1)+1+1-\dfrac2{x+1}+\dfrac1{(x+1)^2}&=3\\ (x+1)^2+\dfrac1{(x+1)^2}-2\left((x+1)+\dfrac1{x+1}\right)+2&=3\\ \left((x+1)+\dfrac1{x+1}\right)^2-2-2\left((x+1)+\dfrac1{x+1}\right)+2&=3\\ \left((x+1)+\dfrac1{x+1}\right)^2-2\left((x+1)+\dfrac1{x+1}\right)-3&=0\\ \left((x+1)+\dfrac1{x+1}+1\right)\left((x+1)+\dfrac1{x+1}-3\right)&=0 \end{aligned}\)

• \((x+1)+\dfrac1{x+1}+1=0\qquad{\color{red}\times(x+1)}\)
  \(\begin{aligned} x^2+2x+1+1+x+1&=0\\ x^2+3x+3&=0 \end{aligned}\)
  D < 0 sehingga tidak ada nilai x yang memenuhi.
  
  
• \((x+1)+\dfrac1{x+1}-3=0\qquad{\color{red}\times(x+1)}\)
  \(\begin{aligned} x^2+2x+1+1-3x-3&=0\\ x^2-x-1&=0\\ x&=\dfrac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}\\[4pt] &=\dfrac{1\pm\sqrt{1+4}}2\\[4pt] &=\dfrac{1\pm\sqrt5}2 \end{aligned}\)

Tinjau bahwa
$\begin{array}{rl}(a\uparrow b)+(b\uparrow a)& ={\displaystyle \frac{a^3}{a-b}}+{\displaystyle \frac{b^3}{b-a}}\\ & ={\displaystyle \frac{a^3}{a-b}}-{\displaystyle \frac{b^3}{a-b}}\\ & ={\displaystyle \frac{a^3-b^3}{a-b}}\\ & =a^2+ab+b^2\end{array}$

$\begin{array}{rl}(10\uparrow 100)+(100\uparrow 10)& ={10}^2+10\cdot 100+{100}^2\\ & =100+1000+10000\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 11100}}}}\end{array}$

Misal m = a², n = b².
\(\begin{aligned} \dfrac{a^2+b^2}2-ab&=1&{\color{red}\times2}\\[4pt] a^2+b^2-2ab&=2\\ (b-a)^2&=2\\ b-a&=\sqrt2\\ b&=a+\sqrt2\\ b^2&=a^2+2a\sqrt2+2 \end{aligned}\)
Perhatikan bahwa b² merupakan bilangan bulat sehingga $a=c\sqrt2$, dengan c merupakan bilangan bulat positif.
\(\begin{aligned} b^2&=(c\sqrt2)^2+2(c\sqrt2)\sqrt2+2\\ n&=2c^2+4c+2\\ n&=2(c+1)^2\leq50\\ c+1&\leq5\\ c&\leq4 \end{aligned}\)