HOTS Zone : Sistem Persamaan Aljabar
Table of Contents
Tipe:
No.
Diketahui x, y, z dan t adalah bilangan real tidak nol dan memenuhi persamaan\({\dfrac1x+\dfrac1y+\dfrac1z=\dfrac1t}\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac1x+\dfrac1y+\dfrac1z&=\dfrac1t\\[4pt]
\dfrac{xy+yz+xz}{xyz}&=\dfrac1t\\[4pt]
xy+yz+xz&=\dfrac{xyz}t
\end{aligned}\)
\(\begin{aligned} x^3+y^3+z^3&=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz\\ 1000^3&=t^3-3\left(\cancel{t}\right)\left(\dfrac{xyz}{\cancel{t}}\right)+3xyz\\ 1000^3&=t^3-\cancel{3xyz}+\cancel{3xyz}\\ t&=10 \end{aligned}\)
\(\begin{aligned} x+y+z+t&=2t\\ &=2(1000)\\ &=\boxed{\boxed{2000}} \end{aligned}\)
\(\begin{aligned} x^3+y^3+z^3&=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz\\ 1000^3&=t^3-3\left(\cancel{t}\right)\left(\dfrac{xyz}{\cancel{t}}\right)+3xyz\\ 1000^3&=t^3-\cancel{3xyz}+\cancel{3xyz}\\ t&=10 \end{aligned}\)
\(\begin{aligned} x+y+z+t&=2t\\ &=2(1000)\\ &=\boxed{\boxed{2000}} \end{aligned}\)
Jadi, nilai dari x + y + z + t adalah 2000.
No.
DiketahuiCarilah nilai dari \(^{\frac1{16}}\log(x+y)\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x^3-3x^2+5x+2016&=2017\\
(x-1)^3+2x&=0\\
(x-1)^3+2(x-1)+2&=0
\end{aligned}\)
Misalp = x − 1
p3 + 2p + 2 = 0
\(\begin{aligned} y^3-3y^2+5y+2017&=2022\\ (y-1)^3+2(y-1)-2&=0 \end{aligned}\)
Misalq = y − 1
q3 + 2q − 2 = 0
\(\begin{aligned} p^3+2p+2&=0\\ q^3+2q-2&=0&\qquad+\\\hline p^3+q^3+2(p+q)&=0\\ (p+q)\left(p^2-pq+q^2\right)+2(p+q)&=0\\ (p+q)\left(p^2-pq+q^2+2\right)&=0\\ p+q&=0\\ x-1+y-1&=0\\ x+y&=2 \end{aligned}\)
Misal
\(\begin{aligned} y^3-3y^2+5y+2017&=2022\\ (y-1)^3+2(y-1)-2&=0 \end{aligned}\)
Misal
\(\begin{aligned} p^3+2p+2&=0\\ q^3+2q-2&=0&\qquad+\\\hline p^3+q^3+2(p+q)&=0\\ (p+q)\left(p^2-pq+q^2\right)+2(p+q)&=0\\ (p+q)\left(p^2-pq+q^2+2\right)&=0\\ p+q&=0\\ x-1+y-1&=0\\ x+y&=2 \end{aligned}\)
Jadi, \(^{\frac1{16}}\log(x+y)=2\).
No.
p, q, dan r adalah tiga bilangan real yang memenuhi persamaanNilai r yang bulat yang memenuhi persamaan tersebut adalah ....
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
r^2-p^2-q^2&=13\\
p^2+q^2&=r^2-13
\end{aligned}\)
\(\begin{aligned} p+q+r&=9\\ (p+q+r)^2&=9^2\\ p^2+q^2+r^2+2(pq+qr+pr)&=81\\ r^2-13+r^2+2(pq+qr+pr)&=81\\ 2r^2+2(pq+qr+pr)&=94\\ pq+qr+pr&=47-r^2 \end{aligned}\)
p, q, dan r merupakan akar-akar dari:
\(\begin{aligned} x^3-9x^2+\left(47-r^2\right)x-10&=0\\ r^3-9r^2+\left(47-r^2\right)r-10&=0\\ r^3-9r^2+47r-r^3-10&=0\\ 9r^2-47r+10&=0\\ (9r-2)(r-5)&=0 \end{aligned}\)
\(r=\dfrac29\) atau r = 5
\(\begin{aligned} p+q+r&=9\\ (p+q+r)^2&=9^2\\ p^2+q^2+r^2+2(pq+qr+pr)&=81\\ r^2-13+r^2+2(pq+qr+pr)&=81\\ 2r^2+2(pq+qr+pr)&=94\\ pq+qr+pr&=47-r^2 \end{aligned}\)
p, q, dan r merupakan akar-akar dari:
\(\begin{aligned} x^3-9x^2+\left(47-r^2\right)x-10&=0\\ r^3-9r^2+\left(47-r^2\right)r-10&=0\\ r^3-9r^2+47r-r^3-10&=0\\ 9r^2-47r+10&=0\\ (9r-2)(r-5)&=0 \end{aligned}\)
\(r=\dfrac29\) atau r = 5
Jadi, nilai r yang bulat yang memenuhi adalah 5.
No.
Diketahui a, b, dan c adalah tiga bilangan real yang memenuhi persamaanmaka
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
a^2-bc&=b^2+ac\\
a^2-b^2&=ac+bc\\
(a+b)(a-b)&=(a+b)c\\
a-b&=c\\
a&=b+c
\end{aligned}\)
\(\begin{aligned} a^2-bc&=7\\ b^2+ac&=7\\ c^2+ab&=7&\qquad+\\\hline a^2+b^2+c^2+ab+ac-bc&=21\\ a^2+b^2+c^2+a(b+c)-bc&=21\\ a^2+b^2+c^2+a(a)-bc&=21\\ a^2+b^2+c^2+a^2-bc&=21\\ a^2+b^2+c^2+7&=21\\ a^2+b^2+c^2&=14 \end{aligned}\)
\(\begin{aligned} a^2-bc&=7\\ b^2+ac&=7\\ c^2+ab&=7&\qquad+\\\hline a^2+b^2+c^2+ab+ac-bc&=21\\ a^2+b^2+c^2+a(b+c)-bc&=21\\ a^2+b^2+c^2+a(a)-bc&=21\\ a^2+b^2+c^2+a^2-bc&=21\\ a^2+b^2+c^2+7&=21\\ a^2+b^2+c^2&=14 \end{aligned}\)
Jadi, a2 + b2 + c2 = 14.
No.
\(\begin{cases} a(b+c-5)=7\\ b(a+c-5)=7\\ a^2+b^2=50 \end{cases}\)Carilah nilai
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
a(b+c-5)&=7\\
ab+ac-5a&=7
\end{aligned}\)
\(\begin{aligned} b(a+c-5)&=7\\ ab+bc-5b&=7 \end{aligned}\)
\(\begin{aligned} ab+ac-5a&=7\\ ab+bc-5b&=7\qquad-\\\hline (a-b)c-5(a-b)&=0\\ (a-b)(c-5)&=0 \end{aligned}\)
\(\begin{aligned} b(a+c-5)&=7\\ ab+bc-5b&=7 \end{aligned}\)
\(\begin{aligned} ab+ac-5a&=7\\ ab+bc-5b&=7\qquad-\\\hline (a-b)c-5(a-b)&=0\\ (a-b)(c-5)&=0 \end{aligned}\)
- 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 a = b = 5
- 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 a2 = 1 maka a = ±1 dan b = ±7
(a, b, c) = {(−1,−7, 5), (1, 7, 5)}
Untuk a2 = 49 makaa = ±7 danb = ±1
(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\(\begin{cases}2a^2+2007a+3=0\\3b^2+2007b+2=0\end{cases}\)
dan
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
2a^2+2007a+3&=0\qquad&\color{red}{\times b}\\
3b^2+2007b+2&=0\qquad&\color{red}{\times a}
\end{aligned}\)
\(\begin{aligned} 2a^2b+2007ab+3b&=0\\ 3ab^2+2007ab+2a&=0\qquad&\color{red}{-}\\\hline 2a^2b-3ab^2+3b-2a&=0\\ ab(2a-3b)+3b-2a&=0\\ (ab-1)(2a-3b)&=0 \end{aligned}\)
\(\begin{aligned} 2a^2b+2007ab+3b&=0\\ 3ab^2+2007ab+2a&=0\qquad&\color{red}{-}\\\hline 2a^2b-3ab^2+3b-2a&=0\\ ab(2a-3b)+3b-2a&=0\\ (ab-1)(2a-3b)&=0 \end{aligned}\)
ab − 1 = 0
ab = 1 TM2a − 3b = 0
\(\begin{aligned} 2a&=3b\\ \dfrac{a}b&=\boxed{\boxed{\dfrac32}} \end{aligned}\)
Jadi, \(\dfrac{a}b=\dfrac32\).
No.
Jika\(\begin{cases}2017a^2+2018a+2019=0\\2019b^2+2018b+2017=0\end{cases}\)
dan
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
2017a^2+2018a+2019&=0\qquad&\color{red}{\times b}\\
2019b^2+2018b+2017&=0\qquad&\color{red}{\times a}
\end{aligned}\)
\(\begin{aligned} 2017a^2b+2018ab+2019b&=0\\ 2019ab^2+2018ab+2017a&=0\qquad&\color{red}{-}\\[-3pt]\hline\\[-12pt] 2017a^2b-2019ab^2+2019b-2017a&=0\\ ab(2017a-2019b)-(2017a-2019b)&=0\\ (ab-1)(2017a-2019b)&=0 \end{aligned}\)
\(\begin{aligned} 2017a^2b+2018ab+2019b&=0\\ 2019ab^2+2018ab+2017a&=0\qquad&\color{red}{-}\\[-3pt]\hline\\[-12pt] 2017a^2b-2019ab^2+2019b-2017a&=0\\ ab(2017a-2019b)-(2017a-2019b)&=0\\ (ab-1)(2017a-2019b)&=0 \end{aligned}\)
ab − 1 = 0
ab = 1 TM2017a − 2019b = 0
\(\begin{aligned} 2017a&=2019b\\ \dfrac{a}b&=\boxed{\boxed{\dfrac{2019}{2017}}} \end{aligned}\)
Jadi, \(\dfrac{a}b=\dfrac{2019}{2017}\).
No.
Diketahui- −18
- −12
- 14
- 18
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \dfrac{2x^2+xy-3y^2}{x-y}&=\dfrac{45}9\\ \dfrac{(x-y)(2x+3y)}{x-y}&=5\\ 2x+3y&=5\qquad(2) \end{aligned}\)
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
JAWAB: A
No.
Jika \({x+y+3\sqrt{x+y}=18}\) dan \({x-y-2\sqrt{x-y}=15}\), makaALTERNATIF PENYELESAIAN
Misal \({a=\sqrt{x+y}\gt0}\) dan \({b=\sqrt{x-y}\gt0}\)
\(\begin{aligned} a^2+3a&=18\\ a^2+3a-18&=0\\ (a+6)(a-3)&=0 \end{aligned}\)
a = −6 (TM) ataua = 3 ⟶ x + y = 9
\(\begin{aligned} b^2-2b&=15\\ b^2-2b-15&=0\\ (b+3)(b-5)&=0 \end{aligned}\)
b = −3 (TM) ataub = 5 ⟶ x − y = 25
\(\begin{aligned} x\cdot y&=\dfrac{(x+y)^2-(x-y)^2}4\\ &=\dfrac{9^2-25^2}4\\ &=\dfrac{9^2-25^2}4\\ &=\boxed{\boxed{-136}} \end{aligned}\)
\(\begin{aligned} a^2+3a&=18\\ a^2+3a-18&=0\\ (a+6)(a-3)&=0 \end{aligned}\)
a = −6 (TM) atau
\(\begin{aligned} b^2-2b&=15\\ b^2-2b-15&=0\\ (b+3)(b-5)&=0 \end{aligned}\)
b = −3 (TM) atau
\(\begin{aligned} x\cdot y&=\dfrac{(x+y)^2-(x-y)^2}4\\ &=\dfrac{9^2-25^2}4\\ &=\dfrac{9^2-25^2}4\\ &=\boxed{\boxed{-136}} \end{aligned}\)
Jadi, x ⋅ y = −136.
No.
DiketahuiALTERNATIF PENYELESAIAN
\(\begin{aligned}
a^2-b^2&=3\\
(a-b)(a+b)&=3\\
(1)(a+b)&=3\\
a+b&=3
\end{aligned}\)
\(\begin{aligned} a-b&=1\\ a+b&=3&\qquad{\color{red}+}\\\hline\\[-12pt] 2a&=4\\ a&=2 \end{aligned}\)
\(\begin{aligned} a-b&=1\\ a+b&=3&\qquad{\color{red}+}\\\hline\\[-12pt] 2a&=4\\ a&=2 \end{aligned}\)
\(\begin{aligned} a^3-b^3&=2^3-1^3\\ &=8-1\\ &=\boxed{\boxed{7}} \end{aligned}\)
Jadi, a3 − b3 = 7.
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