Berikut ini adalah kumpulan soal mengenai Sistem Persamaan tingkat olimpiade. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
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....
\(\eqalign{
\dfrac1x+\dfrac1y+\dfrac1z&=\dfrac1t\\[4pt]
\dfrac{xy+yz+xz}{xyz}&=\dfrac1t\\[4pt]
xy+yz+xz&=\dfrac{xyz}t}\)
\(\eqalign{
x^3+y^3+z^3&=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz\\
1000^3&=t^3-3\left(\cancel{t}\right)\left(\dfrac{xyz}{\cancel{t}}\right)+3xyz\\
1000^3&=t^3-\cancel{3xyz}+\cancel{3xyz}\\
t&=10}\)
\(\eqalign{
x+y+z+t&=2t\\
&=2(1000)\\
&=\boxed{\boxed{2000}}}\)
No. 2
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}}\negmedspace\log(x+y)}
\(\eqalign{
x^3-3x^2+5x+2016&=2017\\
(x-1)^3+2x&=0\\
(x-1)^3+2(x-1)+2&=0
}\)
Misal p=x-1
p^3+2p+2=0
\(\eqalign{
y^3-3y^2+5y+2017&=2022\\
(y-1)^3+2(y-1)-2&=0
}\)
Misal q=y-1
q^3+2q-2=0
\(\eqalign{
p^3+2p+2&=0\\
q^3+2q-2&=0&\qquad+\\\hline
p^3+q^3+2(p+q)&=0\\
(p+q)\left(p^2-pq+q^2\right)+2(p+q)&=0\\
(p+q)\left(p^2-pq+q^2+2\right)&=0\\
p+q&=0\\
x-1+y-1&=0\\
x+y&=2
}\)
No. 3
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 ....
\(\eqalign{
r^2-p^2-q^2&=13\\
p^2+q^2&=r^2-13
}\)
\(\eqalign{
p+q+r&=9\\
(p+q+r)^2&=9^2\\
p^2+q^2+r^2+2(pq+qr+pr)&=81\\
r^2-13+r^2+2(pq+qr+pr)&=81\\
2r^2+2(pq+qr+pr)&=94\\
pq+qr+pr&=47-r^2
}\)
p, q, dan r merupakan akar-akar dari:
\(\eqalign{
x^3-9x^2+\left(47-r^2\right)x-10&=0\\
r^3-9r^2+\left(47-r^2\right)r-10&=0\\
r^3-9r^2+47r-r^3-10&=0\\
9r^2-47r+10&=0\\
(9r-2)(r-5)&=0
}\)
r=\dfrac29 atau r=5
No. 4
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= ....
\(\eqalign{
a^2-bc&=b^2+ac\\
a^2-b^2&=ac+bc\\
(a+b)(a-b)&=(a+b)c\\
a-b&=c\\
a&=b+c
}\)
\(\eqalign{
a^2-bc&=7\\
b^2+ac&=7\\
c^2+ab&=7&\qquad+\\\hline
a^2+b^2+c^2+ab+ac-bc&=21\\
a^2+b^2+c^2+a(b+c)-bc&=21\\
a^2+b^2+c^2+a(a)-bc&=21\\
a^2+b^2+c^2+a^2-bc&=21\\
a^2+b^2+c^2+7&=21\\
a^2+b^2+c^2&=14
}\)
No. 5
Diberikan sistem persamaan
x+3y+2w=4444,
6x+2y=8888,
6z+3w=6666,
Tentukan hasil dari
x+y+z+w
\(\begin{aligned}
x+3y+2w&=4444\\
6x+2y&=8888&\color{red}{:2}\\
6z+3w&=6666&\color{red}{\times\dfrac23}
\end{aligned}\)
\(\begin{aligned}
x+3y+2w&=4444\\
3x+y&=4444\\
4z+2w&=4444&\color{red}{+}\\\hline
4x+4y+4z+4w&=3\cdot4444\\
x+y+z+w&=3\cdot1111\\
&=\boxed{\boxed{3333}}
\end{aligned}\)
No. 6
\begin{cases}
a(b+c-5)=7\\
b(a+c-5)=7\\
a^2+b^2=50
\end{cases}
Carilah nilai
(a,b,c)
\(\begin{aligned}
a(b+c-5)&=7\\
ab+ac-5a&=7
\end{aligned}\)
\(\begin{aligned}
b(a+c-5)&=7\\
ab+bc-5b&=7
\end{aligned}\)
\(\begin{aligned}
ab+ac-5a&=7\\
ab+bc-5b&=7\qquad-\\\hline
(a-b)c-5(a-b)&=0\\
(a-b)(c-5)&=0
\end{aligned}\)
- Untuk a=b
\(\begin{aligned}
a^2+b^2&=50\\
a^2+a^2&=50\\
2a^2&=50\\
a^2&=25\\
a&=\pm5
\end{aligned}\)
- Untuk a=b=5
\begin{aligned}
a(b+c-5)&=7\\
5(5+c-5)&=7\\
5c&=7\\
c&=\dfrac75
\end{aligned}
(a,b,c)=\left(5,5,\dfrac75\right)
- Untuk a=b=-5
\(\begin{aligned}
a(b+c-5)&=7\\
-5(-5+c-5)&=7\\
-5(c-10)&=7\\
-5c+50&=7\\
-5c&=-43\\
c&=\dfrac{43}5
\end{aligned}\)
(a,b,c)=\left(-5,-5,\dfrac{43}5\right)
- Untuk c=5
\(\begin{aligned}
a(b+c-5)&=7\\
a(b+5-5)&=7\\
ab&=7\\
b&=\dfrac7a
\end{aligned}\)
\(\begin{aligned}
a^2+b^2&=50\\
a^2+\left(\dfrac7a\right)^2&=50\\
a^2+\dfrac{49}{a^2}&=50\\
a^4+49&=50a^2\\
a^4-50a^2+49&=0\\
\left(a^2-1\right)\left(a^2-49\right)&=0
\end{aligned}\)
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)\}
No. 7
Diketahui sistem persamaan
{a+3b+2d=8},
{6a+2b=8}, dan
{6c+3d=12}. Maka nilai
{a+b+c+d=}...
\(\begin{aligned}
6a+2b&=8\qquad\color{red}:2\\
3a+b&=4
\end{aligned}\)
\(\begin{aligned}
6c+3d&=12\qquad\color{red}\times\dfrac23\\
4c+2d&=8
\end{aligned}\)
\(\begin{array}{rcrcrcrcl}
a&+&3b&+&&&2d&=&8\\
3a&+&b&&&&&=&4\\
&&&&4c&+&2d&=&8\qquad\color{red}+\\\hline
4a&+&4b&+&4c&+&4d&=&20\\
a&+&b&+&c&+&d&=&\boxed{\boxed{5}}
\end{array}\)
No. 8
Jika begin{cases}2a^2+2007a+3=0\\3b^2+2007b+2=0\end{cases} dan
{ab\neq1}, tentukan
\dfrac{a}b
\(\begin{aligned}
2a^2+2007a+3&=0\qquad&\color{red}{\times b}\\
3b^2+2007b+2&=0\qquad&\color{red}{\times a}
\end{aligned}\)
\(\begin{aligned}
2a^2b+2007ab+3b&=0\\
3ab^2+2007ab+2a&=0\qquad&\color{red}{-}\\[-3pt]\hline\\[-12pt]
2a^2b-3ab^2+3b-2a&=0\\
ab(2a-3b)+3b-2a&=0\\
(ab-1)(2a-3b)&=0
\end{aligned}\)
- ab-1=0
ab=1 TM
- 2a-3b=0
\(\begin{aligned}
2a&=3b\\
\dfrac{a}b&=\boxed{\boxed{\dfrac32}}
\end{aligned}\)
No. 9
Jika \begin{cases}2017a^2+2018a+2019=0\\2019b^2+2018b+2017=0\end{cases} dan
{ab\neq1}, tentukan
\dfrac{a}b
\(\begin{aligned}
2017a^2+2018a+2019&=0\qquad&\color{red}{\times b}\\
2019b^2+2018b+2017&=0\qquad&\color{red}{\times a}
\end{aligned}\)
\(\begin{aligned}
2017a^2b+2018ab+2019b&=0\\
2019ab^2+2018ab+2017a&=0\qquad&\color{red}{-}\\[-3pt]\hline\\[-12pt]
2017a^2b-2019ab^2+2019b-2017a&=0\\
ab(2017a-2019b)-(2017a-2019b)&=0\\
(ab-1)(2017a-2019b)&=0
\end{aligned}\)
- ab-1=0
ab=1 TM
- 2017a-2019b=0
\(\begin{aligned}
2017a&=2019b\\
\dfrac{a}b&=\boxed{\boxed{\dfrac{2019}{2017}}}
\end{aligned}\)
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