HOTS Zone : Bilangan Pecahan

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Tipe:

Standar SBMPTN HOTS

No.

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

misal x=\sqrt2+\dfrac{\sqrt2}{\sqrt2+\dfrac{\sqrt2}{\sqrt2+\cdots}}
\begin{aligned} x&=\sqrt2+\dfrac{\sqrt2}x\\ x^2&=\sqrt2x+\sqrt2\\ x^2-\sqrt2x-\sqrt2&=0\\ x&=\dfrac{\sqrt2+\sqrt{\left(-\sqrt2\right)^2-4(1)(\sqrt2)}}{2(1)}\\ &=\dfrac{\sqrt2+\sqrt{2+4\sqrt2}}2 \end{aligned}

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

No.

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{aligned} \dfrac{983}{466}&=2+\dfrac{51}{466}\\[8pt] &=2+\dfrac1{\dfrac{466}{51}}\\[18pt] &=2+\dfrac1{9+\dfrac7{51}}\\[18pt] &=2+\dfrac1{9+\dfrac1{\dfrac{51}7}}\\[33pt] &=2+\dfrac1{9+\dfrac1{7+\dfrac27}}\\[33pt] &=2+\dfrac1{9+\dfrac1{7+\dfrac1{\dfrac72}}}\\[45pt] &=2+\dfrac1{9+\dfrac1{7+\dfrac1{3+\dfrac12}}}\\[45pt] &=\boxed{2}+\dfrac1{\boxed{9}+\dfrac1{\boxed{7}+\dfrac1{\boxed{3}+\dfrac1{\boxed{1}+1}}}} \end{aligned}

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

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

No.

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{aligned} \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}&=\dfrac{2022^2\times\left(2021^2-2020\right)}{\left(2021+1\right)\times\left(2021-1\right)\times\left(2021+1\right)\times\left(2021^2-2021+1\right)}\times\dfrac{2020^3\times\left(2021^2+2022\right)}{(2021-1)\times\left(2021^2+2021+1\right)}\\ &=\dfrac{2022^2\times\left(2021^2-2020\right)}{\left(2022\right)\times\left(2020\right)\times\left(2022\right)\times\left(2021^2-2020\right)}\times\dfrac{2020^3\times\left(2021^2+2022\right)}{(2020)\times\left(2021^2+2022\right)}\\ &=\boxed{\boxed{2020}} \end{aligned}

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