HOTS Zone : Aljabar [5]
Table of Contents
Tipe:
No.
Berapa banyak n dari 1 hingga 2021 yang mengakibatkanALTERNATIF PENYELESAIAN
Misal FPB dari n2 + 3 dan n + 4 adalah k ≠ 1 .
Misaln + 4 = ka , dan n2 + 3 = kb , dengan a dan b bilangan asli.
n = ka − 4
\(\begin{aligned} n^2+3&=kb\\ (ka-4)^2+3&=kb\\ k^2a^2-8ka+16+3&=kb\\ k^2a^2-8ka+19&=kb \end{aligned}\)
kita lihat bahwak | 19 , sehingga didapat k = 19.
\(\begin{array}{rcccl} 1&\lt&n&\lt&2021\\ 1&\lt&19a-4&\lt&2021\\ 5&\lt&19a&\lt&2025\\ \dfrac5{19}&\lt&a&\lt&\dfrac{2025}{19}\\[3.8pt] 1&\leq&a&\leq&106 \end{array}\)
ada 106 nilai a yang memenuhi, sehingga ada 106 juga nilai n yang memenuhi.
Misal
n = ka − 4
\(\begin{aligned} n^2+3&=kb\\ (ka-4)^2+3&=kb\\ k^2a^2-8ka+16+3&=kb\\ k^2a^2-8ka+19&=kb \end{aligned}\)
kita lihat bahwa
\(\begin{array}{rcccl} 1&\lt&n&\lt&2021\\ 1&\lt&19a-4&\lt&2021\\ 5&\lt&19a&\lt&2025\\ \dfrac5{19}&\lt&a&\lt&\dfrac{2025}{19}\\[3.8pt] 1&\leq&a&\leq&106 \end{array}\)
ada 106 nilai a yang memenuhi, sehingga ada 106 juga nilai n yang memenuhi.
Jadi, ada 106 nilai n dari 1 hingga 2021 yang mengakibatkan n2 + 3 dan n + 4 tidak relatif prima.
No.
Diberikan bilangan real positif x, y, z yang memenuhi sistem persamaan berikutxyz = 1
$x+\dfrac1z=5$
$y+\dfrac1x=29$
$z+\dfrac1y=\dfrac{m}n$
Dengan
ALTERNATIF PENYELESAIAN
Misal $p=z+\dfrac1y=\dfrac{m}n$
\(\begin{aligned} \left(x+\dfrac1z\right)\left(y+\dfrac1x\right)\left(z+\dfrac1y\right)&=x+\dfrac1z+y+\dfrac1x+z+\dfrac1y+xyz+\dfrac1{xyz}\\[4pt] \left(5\right)\left(29\right)\left(p\right)&=5+29+p+1+\dfrac11\\ 145p&=p+36\\ 144p&=36\\ p&=\dfrac{36}{144}\\[4pt] \dfrac{m}n&=\dfrac14 \end{aligned}\)
m = 1
n = 4
m + n = 1 + 4 =5
\(\begin{aligned} \left(x+\dfrac1z\right)\left(y+\dfrac1x\right)\left(z+\dfrac1y\right)&=x+\dfrac1z+y+\dfrac1x+z+\dfrac1y+xyz+\dfrac1{xyz}\\[4pt] \left(5\right)\left(29\right)\left(p\right)&=5+29+p+1+\dfrac11\\ 145p&=p+36\\ 144p&=36\\ p&=\dfrac{36}{144}\\[4pt] \dfrac{m}n&=\dfrac14 \end{aligned}\)
m = 1
n = 4
m + n = 1 + 4 =
Jadi, m + n = 5
No.
DiketahuiALTERNATIF PENYELESAIAN
\(\begin{aligned}
a+\dfrac1{a+2015}&=b-4030+\dfrac1{b-2015}\\[4pt]
a+2015+\dfrac1{a+2015}&=b-2015+\dfrac1{b-2015}
\end{aligned}\)
Misalx = a + 2015, dan y = b − 2015 . Jika x = y, maka |a − b| = 4030, sehingga x ≠ y.
\(\begin{aligned} x+\dfrac1x&=y+\dfrac1y\\[4pt] x-y&=\dfrac1y-\dfrac1x\\[4pt] x-y&=\dfrac{x-y}{xy}\\[4pt] xy&=1\\ (a+2015)(b-2015)&=1\\ ab-2015a+2015b-4060225&=1\\ ab-2015a+2015b&=4060226\\ \dfrac{ab}{2015}-a+b&=\color{blue}\boxed{\boxed{\color{black}\dfrac{4060226}{2015}}} \end{aligned}\)
Misal
\(\begin{aligned} x+\dfrac1x&=y+\dfrac1y\\[4pt] x-y&=\dfrac1y-\dfrac1x\\[4pt] x-y&=\dfrac{x-y}{xy}\\[4pt] xy&=1\\ (a+2015)(b-2015)&=1\\ ab-2015a+2015b-4060225&=1\\ ab-2015a+2015b&=4060226\\ \dfrac{ab}{2015}-a+b&=\color{blue}\boxed{\boxed{\color{black}\dfrac{4060226}{2015}}} \end{aligned}\)
Jadi, $\dfrac{ab}{2015}-a+b=\dfrac{4060226}{2015}$
No.
Hitunglah $$\dfrac{\left(10^4+2^6\right)\left(18^4+2^6\right)\left(26^4+2^6\right)\left(34^4+2^6\right)\left(42^4+2^6\right)}{\left(6^4+2^6\right)\left(14^4+2^6\right)\left(22^4+2^6\right)\left(30^4+2^6\right)\left(38^4+2^6\right)}$$ALTERNATIF PENYELESAIAN
\(\begin{aligned}
x^4+2^6&=x^4+4\cdot2^4\\
&=\left(x^2+2\cdot2^2-2\cdot2\cdot x\right)\left(x^2+2\cdot2^2+2\cdot2\cdot x\right)\\
&=\left(x(x-4)+8\right)\left(x(x+4)+8\right)\\
\end{aligned}\)
\(\begin{aligned} \dfrac{\left(10^4+2^6\right)\left(18^4+2^6\right)\left(26^4+2^6\right)\left(34^4+2^6\right)\left(42^4+2^6\right)}{\left(6^4+2^6\right)\left(14^4+2^6\right)\left(22^4+2^6\right)\left(30^4+2^6\right)\left(38^4+2^6\right)}&=\dfrac{\left(10(10-4)+8\right)\left(10(10+4)+8\right)\left(18(18-4)+8\right)\left(18(18+4)+8\right)\cdots\left(42(42-4)+8\right)\left(42(42+4)+8\right)}{\left(6(6-4)+8\right)\left(6(6+4)+8\right)\left(14(14-4)+8\right)\left(14(14+4)+8\right)\cdots\left(38(38-4)+8\right)\left(38(38+4)+8\right)}\\[4pt] &=\dfrac{\left(10(6)+8\right)\left(10(14)+8\right)\left(18(14)+8\right)\left(18(22)+8\right)\cdots\left(42(38)+8\right)\left(42(46)+8\right)}{\left(6(2)+8\right)\left(6(10)+8\right)\left(14(10)+8\right)\left(14(18)+8\right)\cdots\left(38(34)+8\right)\left(38(42)+8\right)}\\[4pt] &=\dfrac{1940}{20}\\ &=\color{blue}\boxed{\boxed{\color{black}97}} \end{aligned}\)
\(\begin{aligned} \dfrac{\left(10^4+2^6\right)\left(18^4+2^6\right)\left(26^4+2^6\right)\left(34^4+2^6\right)\left(42^4+2^6\right)}{\left(6^4+2^6\right)\left(14^4+2^6\right)\left(22^4+2^6\right)\left(30^4+2^6\right)\left(38^4+2^6\right)}&=\dfrac{\left(10(10-4)+8\right)\left(10(10+4)+8\right)\left(18(18-4)+8\right)\left(18(18+4)+8\right)\cdots\left(42(42-4)+8\right)\left(42(42+4)+8\right)}{\left(6(6-4)+8\right)\left(6(6+4)+8\right)\left(14(14-4)+8\right)\left(14(14+4)+8\right)\cdots\left(38(38-4)+8\right)\left(38(38+4)+8\right)}\\[4pt] &=\dfrac{\left(10(6)+8\right)\left(10(14)+8\right)\left(18(14)+8\right)\left(18(22)+8\right)\cdots\left(42(38)+8\right)\left(42(46)+8\right)}{\left(6(2)+8\right)\left(6(10)+8\right)\left(14(10)+8\right)\left(14(18)+8\right)\cdots\left(38(34)+8\right)\left(38(42)+8\right)}\\[4pt] &=\dfrac{1940}{20}\\ &=\color{blue}\boxed{\boxed{\color{black}97}} \end{aligned}\)
Jadi, $\dfrac{\left(10^4+2^6\right)\left(18^4+2^6\right)\left(26^4+2^6\right)\left(34^4+2^6\right)\left(42^4+2^6\right)}{\left(6^4+2^6\right)\left(14^4+2^6\right)\left(22^4+2^6\right)\left(30^4+2^6\right)\left(38^4+2^6\right)}=97$.
No.
Tentukan banyak bilangan asli n sedemikian sehingga \[\frac{n^{2024}}{n+1}+\frac5{n^2+n}\] bilangan asliALTERNATIF PENYELESAIAN
\(\dfrac{n^{2024}}{n+1}+\dfrac5{n^2+n}=\dfrac{n^{2025}+5}{n^2+n}\)
\(\begin{aligned} n^{2025}+5&=n^{2025}+n^{2024}-n^{2024}+5\\ &=-n^{2024}+5&\pmod{n^2+n}\\ &=-n^{2024}-n^{2023}+n^{2023}+5&\pmod{n^2+n}\\ &=n^{2023}+5&\pmod{n^2+n}\\ &=n+5&\pmod{n^2+n}\\ \end{aligned}\)
agar \(\dfrac{n+5}{n^2+n}\) bilangan asli, maka
\(\begin{aligned} n^2+n&\leq n+5\\ n^2&\leq5 \end{aligned}\)
n = 1 atau n = 2
\(\begin{aligned} n^{2025}+5&=n^{2025}+n^{2024}-n^{2024}+5\\ &=-n^{2024}+5&\pmod{n^2+n}\\ &=-n^{2024}-n^{2023}+n^{2023}+5&\pmod{n^2+n}\\ &=n^{2023}+5&\pmod{n^2+n}\\ &=n+5&\pmod{n^2+n}\\ \end{aligned}\)
agar \(\dfrac{n+5}{n^2+n}\) bilangan asli, maka
\(\begin{aligned} n^2+n&\leq n+5\\ n^2&\leq5 \end{aligned}\)
n = 1
\(\begin{aligned} \dfrac{n+5}{n^2+n}&=\dfrac{1+5}{1^2+1}\\[3.8pt] &=3 \end{aligned}\)n = 2
\(\begin{aligned} \dfrac{n+5}{n^2+n}&=\dfrac{2+5}{2^2+2}\\[3.8pt] &=\dfrac76 \end{aligned}\)
Bukan bilangan bulat
Jadi, banyak bilangan asli n sedemikian sehingga \[\frac{n^{2024}}{n+1}+\frac5{n^2+n}\] bilangan asli ada 1 bilangan.
No.
Pertidaksamaan berikut berlaku untuk semua bilangan bulat positif n. $$\sqrt{n+1}-\sqrt{n}\lt\dfrac1{\sqrt{4n+1}}\lt\sqrt{n}-\sqrt{n-1}$$ Berapa nilai bilangan bulat terbesar yang kurang dari $\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}$?ALTERNATIF PENYELESAIAN
\begin{array}{rcccl}
\displaystyle\sum_{n=1}^{24}\left(\sqrt{n+1}-\sqrt{n}\right)&\lt&\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}&\lt&\displaystyle\sum_{n=1}^{24}\left(\sqrt{n}-\sqrt{n-1}\right)\\[4pt]
\sqrt2-\sqrt1+\sqrt3-\sqrt2+\cdots+\sqrt{25}-\sqrt{24}&\lt&\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}&\lt&\sqrt1-\sqrt0+\sqrt2-\sqrt1+\cdots+\sqrt{24}-\sqrt{23}\\[4pt]
-\sqrt1+\sqrt{25}&\lt&\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}&\lt&-\sqrt0+\sqrt{24}\\[4pt]
4&\lt&\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}&\lt&\sqrt{24}\\[4pt]
4&\lt&\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}&\lt&4,...
\end{array}
Jadi, nilai bilangan bulat terbesar yang kurang dari $\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}$ adalah 4.
No.
Diberikan bilangan real x, y yang memenuhi persamaanALTERNATIF PENYELESAIAN
\(\begin{aligned}
x+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+y&=x+y+\dfrac{x^4+y^4}{x^2y^2}\\[4pt]
&=x+y+\dfrac{(x+y)^4-4xy(x+y)^2+2(xy)^2}{(xy)^2}\\[4pt]
&=4+\dfrac{4^4-4(-2)(4)^2+2(-2)^2}{(-2)^2}\\[4pt]
&=4+\dfrac{256+128+8}{4}\\[4pt]
&=4+\dfrac{392}4\\[4pt]
&=4+98\\
&=\color{blue}\boxed{\boxed{\color{black}102}}
\end{aligned}\)
Jadi, nilai dari $$x+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+y$$ adalah 102.
No.
Diketahui a dan b dua bilangan real nonnegatif yang memenuhi- 49
- 45
- 51
- 41
ALTERNATIF PENYELESAIAN
\(\begin{aligned} 3a+5b&=3(15-2b)+5b\\ &=45-6b+5b\\ &=45-b \end{aligned}\)
Nilai
Jadi, nilai terbesar yang mungkin dari 3a + 5b adalah 45.
JAWAB: B
JAWAB: B
No.
Misalkan x, y bilangan real positif denganALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(\dfrac{x+y}{x-y}\right)^2&=\dfrac{x^2+y^2+2xy}{x^2+y^2-2xy}\\[4pt]
&=\dfrac{\left(\dfrac{545}{272}\right)xy+2xy}{\left(\dfrac{545}{272}\right)xy-2xy}\\[4pt]
&=\dfrac{\left(\dfrac{1089}{272}\right)xy}{\left(\dfrac1{272}\right)xy}\\[4pt]
&=1089\\
\dfrac{x+y}{x-y}&=\sqrt{1089}\\
&=\color{blue}\boxed{\boxed{\color{black}33}}
\end{aligned}\)
Jadi, $\dfrac{x+y}{x-y}=33$.
No.
Jika \(x+y+3\sqrt{x+y}=18\) dan \(x-y-2\sqrt{x-y}=15\), makaALTERNATIF PENYELESAIAN
\(\begin{aligned}
x+y+3\sqrt{x+y}&=18\\
\left(\sqrt{x+y}\right)^2+3\sqrt{x+y}-18&=0\\
\left(\sqrt{x+y}+6\right)\left(\sqrt{x+y}-3\right)&=0\\
\sqrt{x+y}&=3\\
x+y&=9
\end{aligned}\)
\(\begin{aligned} x-y-2\sqrt{x-y}&=15\\ \left(\sqrt{x-y}\right)^2-2\sqrt{x-y}-15&=0\\ \left(\sqrt{x-y}+3\right)\left(\sqrt{x-y}-5\right)&=0\\ \sqrt{x-y}&=5\\ x-y&=25 \end{aligned}\)
\(\begin{aligned} 4xy&=\left(x+y\right)^2-\left(x-y\right)^2\\ &=9^2-25^2\\ &=81-625\\ &=-544\\ xy&=\boxed{\boxed{-138}} \end{aligned}\)
\(\begin{aligned} x-y-2\sqrt{x-y}&=15\\ \left(\sqrt{x-y}\right)^2-2\sqrt{x-y}-15&=0\\ \left(\sqrt{x-y}+3\right)\left(\sqrt{x-y}-5\right)&=0\\ \sqrt{x-y}&=5\\ x-y&=25 \end{aligned}\)
\(\begin{aligned} 4xy&=\left(x+y\right)^2-\left(x-y\right)^2\\ &=9^2-25^2\\ &=81-625\\ &=-544\\ xy&=\boxed{\boxed{-138}} \end{aligned}\)
Jadi, xy = −138.
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