HOTS Zone : Aljabar [2]

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STANDARSNBTHOTS

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No. 11

Diketahui x,y ∈ ℝ, x > 2016 dan x > 2017. Jika $2016\sqrt{(x+2016)(x-2016)}+2017\sqrt{(x+2017)(x-2017)}={\displaystyle \frac{1}{2}}(x^2+y^2)$
maka nilai xy =

1. 4066272
2. 4068289
3. 5750577,011

4. 5756281,95
5. 8132544

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ALTERNATIF PENYELESAIAN

Misal $a=\sqrt{(x+2016)(x-2016)}$

$\begin{array}{rl}{a}^2& =x^2-{2016}^2\\ x^2& =a^2+{2016}^2\end{array}$

Misal $b=\sqrt{(y+2017)(y-2017)}$

$\begin{array}{rl}{b}^2& =y^2-{2017}^2\\ y^2& =b^2+{2017}^2\end{array}$

$\begin{array}{rl}2016a+2017b& ={\displaystyle \frac{1}{2}}(a^2+{2016}^2+b^2+{2017}^2)\\ 2\cdot 2016a+2\cdot 2017b& =a^2+{2016}^2+b^2+{2017}^2\\ a^2-2\cdot 2016a+{2016}^2+b^2-2\cdot 2017b+{2017}^2& =0\\ (a-2016{)}^2+(b-2017{)}^2& =0\end{array}$
didapat a= 2016 dan b= 2017

$\begin{array}{rl}{x}^2& ={2016}^2+{2016}^2\\ & =2\cdot {2016}^2\\ x& =2016\sqrt{2}\end{array}$

$\begin{array}{rl}{y}^2& ={2017}^2+{2017}^2\\ & =2\cdot {2017}^2\\ x& =2017\sqrt{2}\end{array}$

$\begin{array}{rl}xy& =2016\sqrt{2}\cdot 2017\sqrt{2}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8132544}}}}\end{array}$

Jadi, xy = 8132544.
JAWAB: E

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No. 12

Jika $n-{\displaystyle \frac{1}{n}}=x$, maka berapakah $n^2+{\displaystyle \frac{1}{n^2}}$ jika dinyatakan dalam x?

1. x² + 1
2. x + 1
3. x³ + 1

4. x² + 2
5. $\sqrt{x}+\sqrt{1}$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}n-{\displaystyle \frac{1}{n}}& =x\\ {(n-{\displaystyle \frac{1}{n}})}^2& =x^2\\ n^2-2(n)\left({\displaystyle \frac{1}{n}}\right)+{\left({\displaystyle \frac{1}{n}}\right)}^2& =x^2\\ n^2-2+{\displaystyle \frac{1}{n^2}}& =x^2\\ n^2+{\displaystyle \frac{1}{n^2}}& =\boxed{{\displaystyle \boxed{{\displaystyle x^2+2}}}}\end{array}$

Jadi, $n^2+{\displaystyle \frac{1}{n^2}}=x^2+2$.
JAWAB: D

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No. 13

Jika ${\displaystyle \frac{3a+4b}{2a-2b}}=5$ maka tentukan nilai dari ${\displaystyle \frac{a^2+2b^2}{ab}}$

Diktat Pembinaan Olimpiade

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{3a+4b}{2a-2b}}& =5\\ 3a+4b& =10a-10b\\ 7a& =14b\\ {\displaystyle \frac{a}{b}}& =2\end{array}$

$\begin{array}{rl}{\displaystyle \frac{a^2+2b^2}{ab}}& ={\displaystyle \frac{a}{b}}+2{\displaystyle \frac{b}{a}}\\ & =2+2\left({\displaystyle \frac{1}{2}}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$

Jadi, ${\displaystyle \frac{a^2+2b^2}{ab}}=3$.

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No. 14

Nilai dari ${\displaystyle \frac{(1945+2011{)}^2+(2011-1945{)}^2}{{1945}^2+{2011}^2}}$ adalah ....

Diktat Pembinaan Olimpiade

ALTERNATIF PENYELESAIAN

Misal a = 1945 dan b = 2011

$\begin{array}{rl}{\displaystyle \frac{(1945+2011{)}^2+(2011-1945{)}^2}{{1945}^2+{2011}^2}}& ={\displaystyle \frac{(a+b{)}^2+(b-a{)}^2}{a^2+b^2}}\\ & ={\displaystyle \frac{a^2+2ab+b^2+b^2-2ab+a^2}{a^2+b^2}}\\ & ={\displaystyle \frac{2(a^2+b^2)}{a^2+b^2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$

Jadi, ${\displaystyle \frac{(1945+2011{)}^2+(2011-1945{)}^2}{{1945}^2+{2011}^2}}=2.$

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No. 15

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

Matematika Zone Id

ALTERNATIF PENYELESAIAN

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

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

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Didapat ab = 16 dan c = 0

a + b + c = 8 + 0 = 8

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

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No. 16

Jika $x+{\displaystyle \frac{1}{x}}=4$ maka nilai x³ + x⁻³ = ....

Grup Matematika Idhamdaz

ALTERNATIF PEMBAHASAN

$\begin{array}{rl}{(x+{\displaystyle \frac{1}{x}})}^2& =4^2\\ x^2+2+{\displaystyle \frac{1}{x^2}}& =16\\ x^2+{\displaystyle \frac{1}{x^2}}& =14\end{array}$

$\begin{array}{rl}(x^2+{\displaystyle \frac{1}{x^2}})(x+{\displaystyle \frac{1}{x}})& =(14)(4)\\ x^3+x+{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{x^3}}& =56\\ x^3+4+{\displaystyle \frac{1}{x^3}}& =56\\ x^3+{\displaystyle \frac{1}{x^3}}& =\boxed{{\displaystyle \boxed{{\displaystyle 52}}}}\end{array}$

Jadi, x³ + x⁻³ = 52.

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No. 17

Koefisien x² pada ekspansi (x² + 3x − 1)² adalah ....

1. −2
2. −1
3. 2

4. 7
5. 9

ONSB 2023

ALTERNATIF PENYELESAIAN

(x² + 3x − 1)² = (x² + 3x − 1)(x² + 3x − 1)
Kita ambil satu suku dari setiap kurung sedemikian sehingga hasil kalinya mengandung variabel x².
$\begin{array}{rl}\left(x^2\right)(-1)+(3x)(3x)+(-1)\left(x^2\right)& =-x^2+9x^2-x^2\\ & =7x^2\end{array}$

Jadi, koefisien x² pada ekspansi (x² + 3x − 1)² adalah 7.
JAWAB: D

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No. 18

Diketahui x y memenuhi $${\displaystyle \frac{(x-2023)(y-2024)}{(x-2023{)}^2(y-2024{)}^2}}=-{\displaystyle \frac{1}{2}}$$ Nilai yang mungkin bagi x + y adalah ....

Komunitas Pendidik Matematika

ALTERNATIF PENYELESAIAN

Misal a = x − 2023, dan b = y − 2023

$\begin{array}{rl}{\displaystyle \frac{ab}{a^2+b^2}}& =-{\displaystyle \frac{1}{2}}\\ 2ab& =-a^2-b^2\\ a^2+2ab+b^2& =0\\ (a+b{)}^2& =0\\ a+b& =0\\ x-2023+y-2024& =0\\ x+y& =4047\end{array}$

Jadi, nilai yang mungkin bagi x + y adalah 4047.

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No. 19

Jika x dan y bilangan real yang memenuhi ${\displaystyle \frac{x+40}{y}}+{\displaystyle \frac{569}{xy}}={\displaystyle \frac{26-y}{x}}$, tentukan nilai dari xy.

1. −260
2. −270
3. −280

4. −300
5. −400

Grup Matematika Idhamdaz

ALTERNATIF PENYELESAIAN

$\begin{array}{rlr}{\displaystyle \frac{x+40}{y}}+{\displaystyle \frac{569}{xy}}& ={\displaystyle \frac{26-y}{x}}& \text{color}red\times xy\\ x^2+40x+569& =26y-y^2\\ x^2+40x+400+y^2-26y+169& =0\\ (x+20{)}^2+(y-13{)}^2& =0\end{array}$
x = −20 dan y = 13

xy = −20×13 = −260

Jadi, xy = −260.
JAWAB: A

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No. 20

Jika e, b, c adalah bilangan real yang memenuhi |a − b| = 1, |b − c| = 1, |c − a| = 2 dan a × b × c = 60. Nilai dari ${\displaystyle \frac{a}{bc}}+{\displaystyle \frac{b}{ac}}+{\displaystyle \frac{c}{ab}}-{\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{c}}=...$

Grup Matematika Idhamdaz

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{a}{bc}}+{\displaystyle \frac{b}{ac}}+{\displaystyle \frac{c}{ab}}-{\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{c}}& ={\displaystyle \frac{a^2+b^2+c^2-ab-bc-ac}{abc}}\\ & ={\displaystyle \frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{2abc}}\\ & ={\displaystyle \frac{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2}{2\cdot 60}}\\ & ={\displaystyle \frac{1^2+1^2+2^2}{120}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{20}}}}}}\end{array}$

Jadi,
JAWAB:

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