HOTS Zone : Sistem Persamaan Aljabar [2]
Table of Contents
Tipe:
No.
DiketahuiALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(a+b\right)^2&=1^2\\
a^2+b^2+2ab&=1\\
3+2ab&=1\\
2ab&=-2\\
ab&=-1
\end{aligned}\)
\(\begin{aligned} a^3+b^3&=\left(a+b\right)^3-3ab\left(a+b\right)\\ &=\left(1\right)^3-3(-1)\left(1\right)\\ &=1+3\\ &=\boxed{\boxed{4}} \end{aligned}\)
\(\begin{aligned} a^3+b^3&=\left(a+b\right)^3-3ab\left(a+b\right)\\ &=\left(1\right)^3-3(-1)\left(1\right)\\ &=1+3\\ &=\boxed{\boxed{4}} \end{aligned}\)
Jadi, nilai dari a3 + b3 adalah 4.
No.
DiketahuiALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(a+b\right)^2&=3^2\\
a^2+b^2+2ab&=9\\
5+2ab&=9\\
2ab&=4\\
ab&=2
\end{aligned}\)
\(\begin{aligned} a^3+b^3&=\left(a+b\right)^3-3ab\left(a+b\right)\\ &=\left(3\right)^3-3(2)\left(3\right)\\ &=27-18\\ &=\boxed{\boxed{9}} \end{aligned}\)
\(\begin{aligned} a^3+b^3&=\left(a+b\right)^3-3ab\left(a+b\right)\\ &=\left(3\right)^3-3(2)\left(3\right)\\ &=27-18\\ &=\boxed{\boxed{9}} \end{aligned}\)
Jadi, nilai dari a3 + b3 adalah 9.
No.
JikaALTERNATIF PENYELESAIAN
\(\begin{aligned}
2x^2+xy+2y^2&=20\\
2\left(x^2+y^2\right)+xy&=20\\
2(20)+xy&=20\\
40+xy&=20\\
xy&=20
\end{aligned}\)
\(\begin{aligned} \left(x+y\right)^2&=x^2+y^2+2xy\\ &=8+2(20)\\ &=48\\ x+y&=\pm\sqrt{48}\\ &=\pm4\sqrt3 \end{aligned}\)
\(\begin{aligned} \left(x+y\right)^2&=x^2+y^2+2xy\\ &=8+2(20)\\ &=48\\ x+y&=\pm\sqrt{48}\\ &=\pm4\sqrt3 \end{aligned}\)
Jadi, nilai terbesar yang mungkin dari x + y adalah \(4\sqrt3\).
No.
JikaALTERNATIF PENYELESAIAN
\(\begin{aligned}
x^3+y^3&=18\\
(x+y)\left(x^2-xy+y^2\right)&=18\\
(x+y)(6)&=18\\
x+y&=3\\
(x+y)^3&=3^3\\
x^3+y^3+3xy(x+y)&=27\\
18+3xy(3)&=27\\
9xy&=9\\
xy&=1
\end{aligned}\)
\(\begin{aligned}
x^2-xy+y^2&=6\\
x^2+y^2&=6+xy\\
&=6+1\\
&=\boxed{\boxed{7}}
\end{aligned}\)
Jadi, nilai dari x2 + y2 adalah 7.
No.
Tentukan nilai xyz jika \({y+\dfrac1z=z+\dfrac1x=1}\).ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(x+\dfrac1y\right)\left(y+\dfrac1z\right)\left(z+\dfrac1x\right)&=\left(x+\dfrac1y\right)(1)(1)\\
xyz+x+\dfrac1y+y+\dfrac1z+z+\dfrac1x+\dfrac1{xyz}&=x+\dfrac1y\\
xyz+1+1+\dfrac1{xyz}&=0\\
xyz+2+\dfrac1{xyz}&=0
\end{aligned}\)
Misal xyz = a
\(\begin{aligned} a+2+\dfrac1a&=0&\qquad{\color{red}\times a}\\ a^2+2a+1&=0\\ (a+1)^2&=0\\ a&=-1\\ xyz&=\boxed{\boxed{-1}} \end{aligned}\)
Misal xyz = a
\(\begin{aligned} a+2+\dfrac1a&=0&\qquad{\color{red}\times a}\\ a^2+2a+1&=0\\ (a+1)^2&=0\\ a&=-1\\ xyz&=\boxed{\boxed{-1}} \end{aligned}\)
Jadi, xyz = −1.
No.
Jika \({\dfrac{x+2y}6=\dfrac{2y+3z}8=\dfrac{3z+x}{10}}\), maka tentukan nilai \(\dfrac{3yz-3zx-xy}{x^2-2y^2+3z^2}\).ALTERNATIF PENYELESAIAN
CARA 1
\(\begin{aligned}
\dfrac{x+2y}6&=\dfrac{2y+3z}8\\
8x+16y&=12y+18z\\
18z&=8x+4y\\
9z&=4x+2y&\qquad{\color{red}(I)}
\end{aligned}\)
\(\begin{aligned} \dfrac{x+2y}6&=\dfrac{3z+x}{10}\\ 10x+20y&=18z+6x\\ 4x+20y&=18z\\ 2x+10y&=9z\\ 2x+10y&=4x+2y\\ 8y&=2x\\ x&=4y&\qquad{\color{red}(II)} \end{aligned}\)
Substitusikan (II) ke (I)
\(\begin{aligned} 9z&=4(4y)+2y\\ 9z&=18y\\ z&=2y&\qquad{\color{red}(III)} \end{aligned}\)
Substitusikan (II) dan (III) ke persamaan awal
\(\begin{aligned} \dfrac{3yz-3zx-xy}{x^2-2y^2+3z^2}&=\dfrac{3y(2y)-3(2y)(4y)-(4y)y}{(4y)^2-2y^2+3(2y)^2}\\ &=\dfrac{6y^2-24y^2-4y^2}{16y^2-2y^2+12y^2}\\ &=\dfrac{-22y^2}{26y^2}\\ &=\boxed{\boxed{-\dfrac{11}{13}}} \end{aligned}\)
\(\begin{aligned} \dfrac{x+2y}6&=\dfrac{3z+x}{10}\\ 10x+20y&=18z+6x\\ 4x+20y&=18z\\ 2x+10y&=9z\\ 2x+10y&=4x+2y\\ 8y&=2x\\ x&=4y&\qquad{\color{red}(II)} \end{aligned}\)
Substitusikan (II) ke (I)
\(\begin{aligned} 9z&=4(4y)+2y\\ 9z&=18y\\ z&=2y&\qquad{\color{red}(III)} \end{aligned}\)
Substitusikan (II) dan (III) ke persamaan awal
\(\begin{aligned} \dfrac{3yz-3zx-xy}{x^2-2y^2+3z^2}&=\dfrac{3y(2y)-3(2y)(4y)-(4y)y}{(4y)^2-2y^2+3(2y)^2}\\ &=\dfrac{6y^2-24y^2-4y^2}{16y^2-2y^2+12y^2}\\ &=\dfrac{-22y^2}{26y^2}\\ &=\boxed{\boxed{-\dfrac{11}{13}}} \end{aligned}\)
CARA 2
\({\dfrac{x+2y}6=\dfrac{2y+3z}8=\dfrac{3z+x}{10}=k}\)
\(\begin{aligned} x+2y&=6k\\ -2y-3z&=-8k\\ 3z+x&=10k&\qquad{\color{red}+}\\\hline\\[-10pt] 2x&=8k\\ x&=4k&\qquad{\color{red}(I)} \end{aligned}\)
\(\begin{aligned} x+2y&=6k\\ 4k+2y&=6k\\ 2y&=2k\\ y&=k&\qquad{\color{red}(II)} \end{aligned}\)
\(\begin{aligned} 3z+x&=10k\\ 3z+4k&=10k\\ 3z&=6k\\ z&=2k&\qquad{\color{red}(III)} \end{aligned}\)
Substitusikan (I), (II) dan (III) ke persamaan awal
\(\begin{aligned} \dfrac{3yz-3zx-xy}{x^2-2y^2+3z^2}&=\dfrac{3k(2k)-3(2k)(4k)-(4k)k}{(4k)^2-2k^2+3(2k)^2}\\ &=\dfrac{6k^2-24k^2-4k^2}{16k^2-2k^2+12k^2}\\ &=\dfrac{-22k^2}{26k^2}\\ &=\boxed{\boxed{-\dfrac{11}{13}}} \end{aligned}\)
\(\begin{aligned} x+2y&=6k\\ -2y-3z&=-8k\\ 3z+x&=10k&\qquad{\color{red}+}\\\hline\\[-10pt] 2x&=8k\\ x&=4k&\qquad{\color{red}(I)} \end{aligned}\)
\(\begin{aligned} x+2y&=6k\\ 4k+2y&=6k\\ 2y&=2k\\ y&=k&\qquad{\color{red}(II)} \end{aligned}\)
\(\begin{aligned} 3z+x&=10k\\ 3z+4k&=10k\\ 3z&=6k\\ z&=2k&\qquad{\color{red}(III)} \end{aligned}\)
Substitusikan (I), (II) dan (III) ke persamaan awal
\(\begin{aligned} \dfrac{3yz-3zx-xy}{x^2-2y^2+3z^2}&=\dfrac{3k(2k)-3(2k)(4k)-(4k)k}{(4k)^2-2k^2+3(2k)^2}\\ &=\dfrac{6k^2-24k^2-4k^2}{16k^2-2k^2+12k^2}\\ &=\dfrac{-22k^2}{26k^2}\\ &=\boxed{\boxed{-\dfrac{11}{13}}} \end{aligned}\)
Jadi,
JAWAB:
JAWAB:
No.
Diberikan pasangan bilangan real- 1
- 2
- 3
- 4
- 5
ALTERNATIF PENYELESAIAN
Misal p = a + b, dan \(q=\sqrt{2ab}\gt0\).
pq = 2 → p > 0
p2 + q2 = 5
\(\begin{aligned} p^2+2pq+q^2&=4+5\\ (p+q)^2&=9\\ p+q&=3 \end{aligned}\)
didapatp = 1 dan q = 2, atau q = 1 dan p = 2
pq = 2 → p > 0
\(\begin{aligned} p^2+2pq+q^2&=4+5\\ (p+q)^2&=9\\ p+q&=3 \end{aligned}\)
didapat
- p = 1
a + b = 1
\(\begin{aligned} q&=2\\ \sqrt{2ab}&=2\\ ab&=2 \end{aligned}\)
tidak ada nilai a dan b real yang memenuhi.
- p = 2
a + b = 2
\(\begin{aligned} q&=1\\ \sqrt{2ab}&=1\\ ab&=\dfrac12 \end{aligned}\)
a dan b adalah akar dari
\(x^2-2x+\dfrac12=0\)
akar-akarnya adalah(a, b) atau(b, a) , sehinggaa1 = b2 dana2 = b1
\(\begin{aligned} a_1+a_2+b_1+b_2&=2(a_1+b_1)\\ &=2(2)\\ &=4 \end{aligned}\)
Jadi,
JAWAB:
JAWAB:
No.
Nilai minimum p sedemikian sehingga sistem persamaan \begin{cases}a^2+9&=pb\\b^2+9&=pc\\c^2+9&=pa\end{cases} memiliki solusi bilangan real tak negatif a, b, c dapat dinyatakan sebagai $\dfrac{x}y\sqrt{z}$ dengan x dan y bilangan asli yang relatif prima dan z bilangan asli yang tidak habis dibagi bilangan kuadrat selain 1. Nilai dari- 4
- 8
- 12
- 16
- 20
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
a^2+9&=pb\\
b^2+9&=pc\\
c^2+9&=pa&+\\\hline
a^2+9+b^2+9+c^2+9&=pa+pb+pc\\
a^2-6a+9+b^2-6b+9+c^2-6c+9&=pa+pb+pc-6a-6b-6c\\
(a-3)^2+(b-3)^2+(c-3)^2&=(p-6)(a+b+c)
\end{aligned}\)
Karena(a − 3)2 + (b − 3)2 + (c − 3)2 ≥ 0, maka (p − 6)(a + b + c) ≥ 0.
Karena a, b, dan c bilangan real tak negatif, makaa + b + c ≥ 0, sehingga p − 6 ≥ 0 atau p ≥ 6.
Didapat x = 6, y = 1, z = 1.
x + y + z = 6 + 1 + 1 = 8.
Karena
Karena a, b, dan c bilangan real tak negatif, maka
Didapat x = 6, y = 1, z = 1.
Jadi, x + y + z = 8 .
JAWAB: B
JAWAB: B
No.
Apabila x, y, z adalah solusi riil dari sistem persamaan \begin{aligned}x-\sqrt{yz}&=42\\y-\sqrt{zx}&=6\\z-\sqrt{xy}&=-30\end{aligned} berapakah nilai dariALTERNATIF PENYELESAIAN
Kurangi persamaan (1) dengan (2), didapat
\(\begin{aligned} x-y+\sqrt{z}\left(\sqrt{x}-\sqrt{y}\right)&=36\\ \left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)&=36&(4) \end{aligned}\)
Kurangi persamaan (2) dan (3), didapat
$\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)=36....(5)$
Kita lihat bahwa $\sqrt{x}\gt\sqrt{y}\gt\sqrt{z}$
Dari persamaan (4) dan (5), didapat
\(\begin{aligned} \sqrt{x}-\sqrt{y}&=\sqrt{y}-\sqrt{z}\\ \sqrt{x}+\sqrt{z}&=2\sqrt{y}\\ \sqrt{y}&=\dfrac{\sqrt{x}+\sqrt{z}}2 \end{aligned}\)
Lalu, dari persamaan (2):
\(\begin{aligned} \left(\dfrac{\sqrt{x}+\sqrt{z}}2\right)^2-\sqrt{zx}&=6\\[4pt] \left(\dfrac{\sqrt{x}-\sqrt{z}}2\right)^2&=6\\[4pt] \sqrt{x}-\sqrt{z}&=2\sqrt6 \end{aligned}\)
$\sqrt{x}=3\sqrt6$, $\sqrt{y}=2\sqrt6$, $\sqrt{z}=\sqrt6$
\(\begin{aligned} x+y+z&=\left(3\sqrt6\right)^2+\left(2\sqrt6\right)^2+\left(\sqrt6\right)^2\\ &=\color{blue}\boxed{\boxed{\color{black}84}} \end{aligned}\)
\(\begin{aligned} x-y+\sqrt{z}\left(\sqrt{x}-\sqrt{y}\right)&=36\\ \left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)&=36&(4) \end{aligned}\)
Kurangi persamaan (2) dan (3), didapat
$\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)=36....(5)$
Kita lihat bahwa $\sqrt{x}\gt\sqrt{y}\gt\sqrt{z}$
Dari persamaan (4) dan (5), didapat
\(\begin{aligned} \sqrt{x}-\sqrt{y}&=\sqrt{y}-\sqrt{z}\\ \sqrt{x}+\sqrt{z}&=2\sqrt{y}\\ \sqrt{y}&=\dfrac{\sqrt{x}+\sqrt{z}}2 \end{aligned}\)
Lalu, dari persamaan (2):
\(\begin{aligned} \left(\dfrac{\sqrt{x}+\sqrt{z}}2\right)^2-\sqrt{zx}&=6\\[4pt] \left(\dfrac{\sqrt{x}-\sqrt{z}}2\right)^2&=6\\[4pt] \sqrt{x}-\sqrt{z}&=2\sqrt6 \end{aligned}\)
$\sqrt{x}=3\sqrt6$, $\sqrt{y}=2\sqrt6$, $\sqrt{z}=\sqrt6$
\(\begin{aligned} x+y+z&=\left(3\sqrt6\right)^2+\left(2\sqrt6\right)^2+\left(\sqrt6\right)^2\\ &=\color{blue}\boxed{\boxed{\color{black}84}} \end{aligned}\)
Jadi, x + y + z = 84 .
No.
Diberikan bilangan real a, b, c, d yang memenuhi sistem persamaan berikut:a + b + 2ab = 4
b + c + 2bc = 7
c + d + 2cd = 2
Tentukan nilai dari
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
a+b+2ab&=4\\
2a+2b+4ab&=8\\
(2a+1)(2b+1)-1&=8\\
(2a+1)(2b+1)&=9
\end{aligned}\)
Dengan cara yang sama pada persamaan (2) dan (3) didapat:
(2b + 1)(2c + 1) = 15
(2c + 1)(2d + 1) = 5
\(\begin{aligned} (2a+1)(2d+1)&=\dfrac{(2a+1)(2b+1)(2c+1)(2d+1)}{(2b+1)(2c+1)}\\ &=\dfrac{9\cdot5}{15}\\ &=\color{blue}\boxed{\boxed{\color{black}3}} \end{aligned}\)
Dengan cara yang sama pada persamaan (2) dan (3) didapat:
(2b + 1)(2c + 1) = 15
(2c + 1)(2d + 1) = 5
\(\begin{aligned} (2a+1)(2d+1)&=\dfrac{(2a+1)(2b+1)(2c+1)(2d+1)}{(2b+1)(2c+1)}\\ &=\dfrac{9\cdot5}{15}\\ &=\color{blue}\boxed{\boxed{\color{black}3}} \end{aligned}\)
Jadi, (2a + 1)(2d + 1) = 3.
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