HOTS Zone : Sistem Persamaan

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Tipe:

Standar SBMPTN HOTS

No.

Diketahui x, y, z dan t adalah bilangan real tidak nol dan memenuhi persamaan

{x+y+z=t}
{\dfrac1x+\dfrac1y+\dfrac1z=\dfrac1t}
{x^3+y^3+z^3=1000^3}

Nilai dari {x+y+z+t} adalah....

$\begin{array}{rl}{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}& ={\displaystyle \frac{1}{t}}\\ {\displaystyle \frac{xy+yz+xz}{xyz}}& ={\displaystyle \frac{1}{t}}\\ xy+yz+xz& ={\displaystyle \frac{xyz}{t}}\end{array}$

$\begin{array}{rl}{x}^3+y^3+z^3& =(x+y+z{)}^3-3(x+y+z)(xy+yz+xz)+3xyz\\ {1000}^3& =t^3-3\left(\cancel{t}\right)\left({\displaystyle \frac{xyz}{\cancel{t}}}\right)+3xyz\\ {1000}^3& =t^3-\cancel{3xyz}+\cancel{3xyz}\\ t& =10\end{array}$
$\begin{array}{rl}x+y+z+t& =2t\\ & =2(1000)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2000}}}}\end{array}$

Jadi, nilai dari x+y+z+t adalah 2000.

No.

Diketahui (x,y) memenuhi dua persamaan :
x^3-3x^2+5x+2016=2017
y^3-3y^2+5y+2017=2022
Carilah nilai dari ^{\frac1{16}}\log(x+y)

$\begin{array}{rl}{x}^3-3x^2+5x+2016& =2017\\ (x-1{)}^3+2x& =0\\ (x-1{)}^3+2(x-1)+2& =0\end{array}$
Misal p=x-1
p^3+2p+2=0

$\begin{array}{rl}{y}^3-3y^2+5y+2017& =2022\\ (y-1{)}^3+2(y-1)-2& =0\end{array}$
Misal q=y-1
q^3+2q-2=0

$\begin{array}{rl}{p}^3+2p+2& =0\\ q^3+2q-2& =0& +\\ p^3+q^3+2(p+q)& =0\\ (p+q)(p^2-pq+q^2)+2(p+q)& =0\\ (p+q)(p^2-pq+q^2+2)& =0\\ p+q& =0\\ x-1+y-1& =0\\ x+y& =2\end{array}$

Jadi, ^{\frac1{16}}\log(x+y)=2.

No.

p, q, dan r adalah tiga bilangan real yang memenuhi persamaan
p+q+r=9
pqr=10
r^2-p^2-q^2=13
Nilai r yang bulat yang memenuhi persamaan tersebut adalah ....

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$\begin{array}{rl}{r}^2-p^2-q^2& =13\\ p^2+q^2& =r^2-13\end{array}$

$\begin{array}{rl}p+q+r& =9\\ (p+q+r{)}^2& =9^2\\ p^2+q^2+r^2+2(pq+qr+pr)& =81\\ r^2-13+r^2+2(pq+qr+pr)& =81\\ 2r^2+2(pq+qr+pr)& =94\\ pq+qr+pr& =47-r^2\end{array}$

p, q, dan r merupakan akar-akar dari:

$\begin{array}{rl}{x}^3-9x^2+(47-r^2)x-10& =0\\ r^3-9r^2+(47-r^2)r-10& =0\\ r^3-9r^2+47r-r^3-10& =0\\ 9r^2-47r+10& =0\\ (9r-2)(r-5)& =0\end{array}$

r=\dfrac29 atau r=5

Jadi, nilai r yang bulat yang memenuhi adalah 5.

No.

Diketahui a, b, dan c adalah tiga bilangan real yang memenuhi persamaan
a^2-bc=7
b^2+ac=7
c^2+ab=7
maka a^2+b^2+c^2= ....

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$\begin{array}{rl}{a}^2-bc& =b^2+ac\\ a^2-b^2& =ac+bc\\ (a+b)(a-b)& =(a+b)c\\ a-b& =c\\ a& =b+c\end{array}$

$\begin{array}{rl}{a}^2-bc& =7\\ b^2+ac& =7\\ c^2+ab& =7& +\\ a^2+b^2+c^2+ab+ac-bc& =21\\ a^2+b^2+c^2+a(b+c)-bc& =21\\ a^2+b^2+c^2+a(a)-bc& =21\\ a^2+b^2+c^2+a^2-bc& =21\\ a^2+b^2+c^2+7& =21\\ a^2+b^2+c^2& =14\end{array}$

Jadi, a^2+b^2+c^2=14.

No.

$$\{\begin{array}{l}a(b+c-5)=7\\ b(a+c-5)=7\\ a^2+b^2=50\end{array}$$ Carilah nilai (a,b,c)

$$\begin{array}{rl}a(b+c-5)& =7\\ ab+ac-5a& =7\end{array}$$

$$\begin{array}{rl}b(a+c-5)& =7\\ ab+bc-5b& =7\end{array}$$

$$\begin{array}{rl}ab+ac-5a& =7\\ ab+bc-5b& =7-\\ (a-b)c-5(a-b)& =0\\ (a-b)(c-5)& =0\end{array}$$

• Untuk a=b
  $$\begin{array}{rl}{a}^2+b^2& =50\\ a^2+a^2& =50\\ 2a^2& =50\\ a^2& =25\\ a& =\pm 5\end{array}$$
  
  • Untuk a=b=5
    $$\begin{array}{rl}a(b+c-5)& =7\\ 5(5+c-5)& =7\\ 5c& =7\\ c& ={\displaystyle \frac{7}{5}}\end{array}$$
    (a,b,c)=\left(5,5,\dfrac75\right)
    
  • Untuk a=b=-5
    $$\begin{array}{rl}a(b+c-5)& =7\\ -5(-5+c-5)& =7\\ -5(c-10)& =7\\ -5c+50& =7\\ -5c& =-43\\ c& ={\displaystyle \frac{43}{5}}\end{array}$$
    (a,b,c)=\left(-5,-5,\dfrac{43}5\right)
    
• Untuk c=5
  $$\begin{array}{rl}a(b+c-5)& =7\\ a(b+5-5)& =7\\ ab& =7\\ b& ={\displaystyle \frac{7}{a}}\end{array}$$
  
  $$\begin{array}{rl}{a}^2+b^2& =50\\ a^2+{\left({\displaystyle \frac{7}{a}}\right)}^2& =50\\ a^2+{\displaystyle \frac{49}{a^2}}& =50\\ a^4+49& =50a^2\\ a^4-50a^2+49& =0\\ (a^2-1)(a^2-49)& =0\end{array}$$
  Untuk a^2=1 maka a=\pm1 dan b=\pm7
  (a,b,c)=\{(-1,-7,5),(1,7,5)\}
  Untuk a^2=49 maka a=\pm7 dan b=\pm1
  (a,b,c)=\{(-7,-1,5),(7,1,5)\}
  

Jadi, (a,b,c)=\{\left(5,5,\dfrac75\right),\left(-5,-5,\dfrac{43}5\right),(-1,-7,5),(1,7,5),(-7,-1,5),(7,1,5)\}.

No.

Jika $$\{\begin{array}{l}2a^2+2007a+3=0\\ 3b^2+2007b+2=0\end{array}$$ dan {ab\neq1}, tentukan \dfrac{a}b

$$\begin{array}{rlr}2a^2+2007a+3& =0& \times b\\ 3b^2+2007b+2& =0& \times a\end{array}$$

$$\begin{array}{rl}2a^{2}b+2007ab+3b& =0\\ 3a{b}^2+2007ab+2a& =0& -\\ \\ 2a^{2}b-3a{b}^2+3b-2a& =0\\ ab(2a-3b)+3b-2a& =0\\ (ab-1)(2a-3b)& =0\end{array}$$

• ab-1=0
  ab=1 TM
• 2a-3b=0
  $$\begin{array}{rl}2a& =3b\\ {\displaystyle \frac{a}{b}}& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{2}}}}}}\end{array}$$

Jadi, \dfrac{a}b=\dfrac32.

No.

Jika $$\{\begin{array}{l}2017a^2+2018a+2019=0\\ 2019b^2+2018b+2017=0\end{array}$$ dan {ab\neq1}, tentukan \dfrac{a}b

$$\begin{array}{rlr}2017a^2+2018a+2019& =0& \times b\\ 2019b^2+2018b+2017& =0& \times a\end{array}$$

$$\begin{array}{rl}2017a^{2}b+2018ab+2019b& =0\\ 2019a{b}^2+2018ab+2017a& =0& -\\ \\ 2017a^{2}b-2019a{b}^2+2019b-2017a& =0\\ ab(2017a-2019b)-(2017a-2019b)& =0\\ (ab-1)(2017a-2019b)& =0\end{array}$$

• ab-1=0
  ab=1 TM
• 2017a-2019b=0
  $$\begin{array}{rl}2017a& =2019b\\ {\displaystyle \frac{a}{b}}& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2019}{2017}}}}}}\end{array}$$

Jadi, \dfrac{a}b=\dfrac{2019}{2017}.

No.

Diketahui {x-y=9} dan {2x^2+xy-3y^2=45}. Jika a dan b adalah bilangan bulat terbesar yang kurang dari atau sama dengan x dan y, maka hasil kali a dan b adalah ...

1. -18
2. -12

3. 14
4. 18

KSP SMP 2020

x-y=9\qquad(1) $\begin{array}{rl}{\displaystyle \frac{2x^2+xy-3y^2}{x-y}}& ={\displaystyle \frac{45}{9}}\\ {\displaystyle \frac{(x-y)(2x+3y)}{x-y}}& =5\\ 2x+3y& =5(2)\end{array}$

Dari persamaan (1) dan (2) didapat x=\dfrac{32}5=6{,}4 dan y=-\dfrac{13}5=-2{,}6

a=\lfloor x\rfloor=\lfloor6{,}4\rfloor=6
b=\lfloor x\rfloor=\lfloor-2{,}6\rfloor=-3

a\cdot b=6\cdot(-3)=\boxed{\boxed{-18}}

Jadi, hasil kali a dan b adalah -18. JAWAB: A

No.

Jika {x+y+3\sqrt{x+y}=18} dan {x-y-2\sqrt{x-y}=15}, maka x\cdot y= ....

Diktat Pembinaan Olimpiade

Misal {a=\sqrt{x+y}\gt0} dan {b=\sqrt{x-y}\gt0}

$$\begin{array}{rl}{a}^2+3a& =18\\ a^2+3a-18& =0\\ (a+6)(a-3)& =0\end{array}$$
a=-6 (TM atau {a=3\rightarrow x+y=9}

$$\begin{array}{rl}{b}^2-2b& =15\\ b^2-2b-15& =0\\ (b+3)(b-5)& =0\end{array}$$
b=-3 (TM) atau b=5\rightarrow x-y=25

$$\begin{array}{rl}x\cdot y& ={\displaystyle \frac{(x+y{)}^2-(x-y{)}^2}{4}}\\ & ={\displaystyle \frac{9^2-{25}^2}{4}}\\ & ={\displaystyle \frac{9^2-{25}^2}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -136}}}}\end{array}$$

Jadi, x\cdot y=-136.