HOTS Zone : Persamaan Aljabar

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Berikut ini adalah kumpulan soal mengenai Persamaan. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

STANDARSNBTHOTS

No. 1

Misalkan x dan y adalah dua bilangan riil yang memenuhi $\dfrac{x^{2}+y^{2}}{x^{2}-y^{2}}+\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}=3$. Apabila nilai dari $\dfrac{x^{8}+y^{8}}{x^{8}-y^{8}}+\dfrac{x^{8}-y^{8}}{x^{8}+y^{8}}$ dapat dinyatakan sebagai pecahan sederhana $\dfrac{a}{b}$, dengan a dan b adalah dua bilangan asli, tentukan nilai dari a + b.

KTOM September 2020

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{x^2+y^2}{x^2-y^2}}+{\displaystyle \frac{x^2-y^2}{x^2+y^2}}& =3\\ {\displaystyle \frac{{(x^2+y^2)}^2+{(x^2-y^2)}^2}{(x^2+y^2)(x^2-y^2)}}& =3\\ {\displaystyle \frac{x^4+2x^2{y}^2+y^4+x^4-2x^2{y}^2+y^4}{x^4-y^4}}& =3\\ {\displaystyle \frac{2x^4+2y^4}{x^4-y^4}}& =3\\ 2x^4+2y^4& =3x^4-3y^4\\ 5y^4& =x^4\\ x^8& =25y^8\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{x^8+y^8}{x^8-y^8}}+{\displaystyle \frac{x^8-y^8}{x^8+y^8}}& ={\displaystyle \frac{25y^8+y^8}{25y^8-y^8}}+{\displaystyle \frac{25y^8-y^8}{25y^8+y^8}}\\ & ={\displaystyle \frac{26y^8}{24y^8}}+{\displaystyle \frac{24y^8}{26y^8}}\\ & ={\displaystyle \frac{13}{12}}+{\displaystyle \frac{12}{13}}\\ & ={\displaystyle \frac{169+144}{156}}\\ & ={\displaystyle \frac{313}{156}}\end{array}$$ a = 313, b = 156

a + b = 313 + 156 = 469469

Jadi, a + b = 469.

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No. 2

Selesaikan persamaan $x^2+\dfrac{x^2}{(x+1)^2}=3$

Canadian MO 1992

ALTERNATIF PENYELESAIAN

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

• p + 3 = 0
  $\begin{array}{rl}p& =-3\\ {\displaystyle \frac{x^2}{x+1}}& =-3\\ x^2& =-3x-3\\ x^2+3x+3& =0\end{array}$
  
  $\begin{array}{rl}D& =3^2-4(1)(3)\\ & =9-12\\ & =-3\end{array}$
  Karena D < 0 sehingga persamaan di atas tidak ada solusi real.
• p − 1 = 0
  $\begin{array}{rl}p& =1\\ {\displaystyle \frac{x^2}{x+1}}& =1\\ x^2& =x+1\\ x^2-x-1& =0\\ x& ={\displaystyle \frac{-(-1)\pm \sqrt{(-1{)}^2-4(1)(-1)}}{2(1)}}\\ & ={\displaystyle \frac{1\pm \sqrt{1+4}}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1\pm \sqrt{5}}{2}}}}}}\end{array}$

Jadi, $x=\dfrac{1\pm\sqrt5}2$.

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No. 3

Hasil perkalian dari nilai-nilai x yang memenuhi
$\dfrac{x^2}{10000}=\dfrac{10000}{x^{2\left({^{10}\negmedspace\log x}\right)-8}}$
adalah

1. 10²
2. 10³
3. 10⁴

4. 10⁵
5. 10⁷

Grup Telegram

ALTERNATIF PENYELESAIAN

x > 0

$\begin{array}{rl}{\displaystyle \frac{x^2}{10000}}& ={\displaystyle \frac{{10}^5}{x^{2\left({}^{10}\log x\right)-8}}}\\ {\displaystyle \frac{x^2}{10000}}& ={\displaystyle \frac{{10}^5}{x^{2\log x-8}}}\\ x^2\cdot x^{2\log x-8}& ={10}^5\cdot {10}^5\\ x^{2+2\log x-8}& ={10}^{5+5}\\ x^{2\log x-6}& ={10}^{10}\\ x^{\frac{2\log x-6}{2}}& ={10}^{\frac{10}{2}}\\ x^{\log x-3}& ={10}^5\\ \log \left(x^{\log x-3}\right)& =\log {10}^5\\ (\log x-3)\log x& =5\end{array}$
Misal log x = p
$\begin{array}{rl}(p-3)p& =5\\ p^2-3p-5& =0\end{array}$

$\begin{array}{rl}{p}_1+p_2& ={\displaystyle \frac{-(-3)}{1}}\\ \log x_1+\log x_2& =3\\ \log (x_1\cdot x_2)& =3\\ x_1\cdot x_2& =\boxed{{\displaystyle \boxed{{\displaystyle {10}^3}}}}\end{array}$

Jadi, hasil perkalian dari nilai-nilai x yang memenuhi
$\dfrac{x^2}{10000}=\dfrac{10000}{x^{2\left({^{10}\negmedspace\log x}\right)-8}}$

Related: loading

adalah 10³.
JAWAB: B

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No. 4

Bilangan-bilangan real a, b, dan c memenuhi sistem persamaan a + b = 8 dan ab = c² + 16. Hasil dari a + b + c = ....

1. 4
2. 5
3. 6

4. 7
5. 8

SUMBER

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}ab& \le {\displaystyle \frac{(a+b{)}^2}{4}}\\ & \le {\displaystyle \frac{8^2}{4}}\\ & \le 16\end{array}$

$\begin{array}{rl}{c}^2+16& \ge 16\\ ab& \ge 16\end{array}$
Didapat ab = 16 dan c = 0

a + b + c = 8 + 0 = 8

Jadi, a + b + c = 8.
JAWAB: E

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