HOTS Zone : Persamaan

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

Tipe:

Standar SBMPTN HOTS

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

$$\begin{array}{rl}{\displaystyle \frac{x^2+y^2}{x^2-y^2}}+{\displaystyle \frac{x^2-y^2}{x^2+y^2}}& =3\\ {\displaystyle \frac{{(x^2+y^2)}^2+{(x^2-y^2)}^2}{(x^2+y^2)(x^2-y^2)}}& =3\\ {\displaystyle \frac{x^4+2x^2{y}^2+y^4+x^4-2x^2{y}^2+y^4}{x^4-y^4}}& =3\\ {\displaystyle \frac{2x^4+2y^4}{x^4-y^4}}& =3\\ 2x^4+2y^4& =3x^4-3y^4\\ 5y^4& =x^4\\ x^8& =25y^8\end{array}$$

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

a+b=313+156=\boxed{\boxed{469}}

Jadi, {a+b=469}. Butuh penjelasan dalam video? komen di bawah dan sebutkan nomor berapa yang ingin dibuatkan video penjelasannya.

No. 2

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

Canadian MO 1992

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

$$\begin{array}{rl}p+3& =0\\ p& =-3\\ {\displaystyle \frac{x^2}{x+1}}& =-3\\ x^2& =-3x-3\\ x^2+3x+3& =0\end{array}$$ $$\begin{array}{rl}D& =3^2-4(1)(3)\\ & =9-12\\ & =-3\end{array}$$ Karena D\lt0 sehingga persamaan di atas tidak ada solusi real.

$$\begin{array}{rl}p-1& =0\\ p& =1\\ {\displaystyle \frac{x^2}{x+1}}& =1\\ x^2& =x+1\\ x^2-x-1& =0\\ x& ={\displaystyle \frac{-(-1)\pm \sqrt{(-1{)}^2-4(1)(-1)}}{2(1)}}\\ & ={\displaystyle \frac{1\pm \sqrt{1+4}}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1\pm \sqrt{5}}{2}}}}}}\end{array}$$

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

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. 10^2
2. 10^3
3. 10^4

4. 10^5
5. 10^7

Grup Telegram

x\gt0

$$\begin{array}{rl}{\displaystyle \frac{x^2}{10000}}& ={\displaystyle \frac{{10}^5}{x^{2\left({}^{10}\log x\right)-8}}}\\ {\displaystyle \frac{x^2}{10000}}& ={\displaystyle \frac{{10}^5}{x^{2\log x-8}}}\\ x^2\cdot x^{2\log x-8}& ={10}^5\cdot {10}^5\\ x^{2+2\log x-8}& ={10}^{5+5}\\ x^{2\log x-6}& ={10}^{10}\\ x^{\frac{2\log x-6}{2}}& ={10}^{\frac{10}{2}}\\ x^{\log x-3}& ={10}^5\\ \log \left(x^{\log x-3}\right)& =\log {10}^5\\ (\log x-3)\log x& =5\end{array}$$
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}} adalah 10^3. JAWAB: B

No. 4

Bilangan-bilangan real a, b, dan c memenuhi sistem persamaan {a+b=8} dan {ab=c^2+16}. Hasil dari {a+b+c=} ....

1. 4
2. 5
3. 6

4. 7
5. 8

SUMBER

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

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

Didapat {ab=16} dan {c=0}

a+b+c=8+0=\boxed{\boxed{8}}

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