HOTS Zone : Bilangan Pecahan

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Tipe:

Standar SBMPTN HOTS

No. 1

\sqrt2+\dfrac{\sqrt2}{\sqrt2+\dfrac{\sqrt2}{\sqrt2+\cdots}}= ....

misal x=\sqrt2+\dfrac{\sqrt2}{\sqrt2+\dfrac{\sqrt2}{\sqrt2+\cdots}}
$$\begin{array}{rl}x& =\sqrt{2}+{\displaystyle \frac{\sqrt{2}}{x}}\\ x^2& =\sqrt{2}x+\sqrt{2}\\ x^2-\sqrt{2}x-\sqrt{2}& =0\\ x& ={\displaystyle \frac{\sqrt{2}+\sqrt{{(-\sqrt{2})}^2-4(1)(\sqrt{2})}}{2(1)}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{2+4\sqrt{2}}}{2}}\end{array}$$

Jadi, \sqrt2+\dfrac{\sqrt2}{\sqrt2+\dfrac{\sqrt2}{\sqrt2+\cdots}}=\dfrac{\sqrt2+\sqrt{2+4\sqrt2}}2.

No. 2

Jika {\dfrac{983}{466}=a+\dfrac1{b+\dfrac1{c+\dfrac1{d+\dfrac1{e+1}}}}}, dimana a, b, c, d, dan e adalah bilangan bulat positif, maka nilai dari {a\cdot b\cdot c\cdot d\cdot e} adalah

1. 189
2. 126
   
3. 252
   

4. 233
5. 378
   

$$\begin{array}{rl}{\displaystyle \frac{983}{466}}& =2+{\displaystyle \frac{51}{466}}\\ & =2+{\displaystyle \frac{1}{{\displaystyle \frac{466}{51}}}}\\ & =2+{\displaystyle \frac{1}{9+{\displaystyle \frac{7}{51}}}}\\ & =2+{\displaystyle \frac{1}{9+{\displaystyle \frac{1}{{\displaystyle \frac{51}{7}}}}}}\\ & =2+{\displaystyle \frac{1}{9+{\displaystyle \frac{1}{7+{\displaystyle \frac{2}{7}}}}}}\\ & =2+{\displaystyle \frac{1}{9+{\displaystyle \frac{1}{7+{\displaystyle \frac{1}{{\displaystyle \frac{7}{2}}}}}}}}\\ & =2+{\displaystyle \frac{1}{9+{\displaystyle \frac{1}{7+{\displaystyle \frac{1}{3+{\displaystyle \frac{1}{2}}}}}}}}\\ & =\boxed{{\displaystyle 2}}+{\displaystyle \frac{1}{\boxed{{\displaystyle 9}}+{\displaystyle \frac{1}{\boxed{{\displaystyle 7}}+{\displaystyle \frac{1}{\boxed{{\displaystyle 3}}+{\displaystyle \frac{1}{\boxed{{\displaystyle 1}}+1}}}}}}}}\end{array}$$

2\cdot9\cdot7\cdot3\cdot1=\boxed{\boxed{378}}

Jadi, {a\cdot b\cdot c\cdot d\cdot e=378}. JAWAB: E

No. 3

Nilai dari {\dfrac{2022^2\times\left(2021^2-2020\right)}{\left(2021^2-1\right)\times\left(2021^3+1\right)}\times\dfrac{2020^3\times\left(2021^2+2022\right)}{2021^3-1}} adalah ....

1. 1
2. 2020
3. 2021

4. 2022
5. 2023

C S C POSI 2021

$$\begin{array}{rl}{\displaystyle \frac{{2022}^2\times ({2021}^2-2020)}{({2021}^2-1)\times ({2021}^3+1)}}\times {\displaystyle \frac{{2020}^3\times ({2021}^2+2022)}{{2021}^3-1}}& ={\displaystyle \frac{{2022}^2\times ({2021}^2-2020)}{(2021+1)\times (2021-1)\times (2021+1)\times ({2021}^2-2021+1)}}\times {\displaystyle \frac{{2020}^3\times ({2021}^2+2022)}{(2021-1)\times ({2021}^2+2021+1)}}\\ & ={\displaystyle \frac{{2022}^2\times ({2021}^2-2020)}{\left(2022\right)\times \left(2020\right)\times \left(2022\right)\times ({2021}^2-2020)}}\times {\displaystyle \frac{{2020}^3\times ({2021}^2+2022)}{(2020)\times ({2021}^2+2022)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2020}}}}\end{array}$$

Jadi, \dfrac{2022^2\times\left(2021^2-2020\right)}{\left(2021^2-1\right)\times\left(2021^3+1\right)}\times\dfrac{2020^3\times\left(2021^2+2022\right)}{2021^3-1}=2020. JAWAB: B