HOTS Zone : Aljabar [6]
Table of Contents
Tipe
No.
Selesaikan persamaan \(x^2+\dfrac{x^2}{(x+1)^2}=3\), untukALTERNATIF PENYELESAIAN
\(\begin{aligned}
x^2+\dfrac{x^2}{(x+1)^2}&=3\\
(x+1-1)^2+\dfrac{(x+1-1)^2}{(x+1)^2}&=3\\
(x+1)^2-2(x+1)+1+\dfrac{(x+1)^2-2(x+1)+1}{(x+1)^2}&=3\\
(x+1)^2-2(x+1)+1+1-\dfrac2{x+1}+\dfrac1{(x+1)^2}&=3\\
(x+1)^2+\dfrac1{(x+1)^2}-2\left((x+1)+\dfrac1{x+1}\right)+2&=3\\
\left((x+1)+\dfrac1{x+1}\right)^2-2-2\left((x+1)+\dfrac1{x+1}\right)+2&=3\\
\left((x+1)+\dfrac1{x+1}\right)^2-2\left((x+1)+\dfrac1{x+1}\right)-3&=0\\
\left((x+1)+\dfrac1{x+1}+1\right)\left((x+1)+\dfrac1{x+1}-3\right)&=0
\end{aligned}\)
- \((x+1)+\dfrac1{x+1}+1=0\qquad{\color{red}\times(x+1)}\)
\(\begin{aligned} x^2+2x+1+1+x+1&=0\\ x^2+3x+3&=0 \end{aligned}\)
D < 0 sehingga tidak ada nilai x yang memenuhi.
- \((x+1)+\dfrac1{x+1}-3=0\qquad{\color{red}\times(x+1)}\)
\(\begin{aligned} x^2+2x+1+1-3x-3&=0\\ x^2-x-1&=0\\ x&=\dfrac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}\\[4pt] &=\dfrac{1\pm\sqrt{1+4}}2\\[4pt] &=\dfrac{1\pm\sqrt5}2 \end{aligned}\)
Jadi, \(x=\dfrac{1+\sqrt5}2\) atau \(x=\dfrac{1-\sqrt5}2\).
No.
Definisikan sebuah operasi ↑ di mana untuk setiap bilangan real- 9100
- 10110
- 9010
- 11100
ALTERNATIF PENYELESAIAN
Tinjau bahwa
Jadi, (10 ↑ 100) + (100 ↑ 10) = 11100.
JAWAB: D
JAWAB: D
No.
$$x^2+3+\dfrac{49}{x^2+3}=14$$ Tentukan nilai x.ALTERNATIF PENYELESAIAN
CARA 1
Misal\(\begin{aligned} y+\dfrac{49}y&=14\\ y^2+49&=14y\\ y^2-14y+49&=0\\ (y-7)^2&=0\\ y&=7\\ x^2+3&=7\\ x^2&=4\\ x&=\pm2 \end{aligned}\)
CARA 2
$x^2+3+\dfrac{7^2}{x^2+3}=2\cdot7$\(\begin{aligned} x^2+3&=7\\ x^2&=4\\ x&=\pm2 \end{aligned}\)
Jadi, x = −2 atau x = 2.
No.
Banyaknya pasangan bilangan bulat positif m dan n dengan- 1
- 2
- 3
- 4
- 6
ALTERNATIF PENYELESAIAN
Misal m = a2, n = b2 .
\(\begin{aligned} \dfrac{a^2+b^2}2-ab&=1&{\color{red}\times2}\\[4pt] a^2+b^2-2ab&=2\\ (b-a)^2&=2\\ b-a&=\sqrt2\\ b&=a+\sqrt2\\ b^2&=a^2+2a\sqrt2+2 \end{aligned}\)
Perhatikan bahwab2 merupakan bilangan bulat sehingga $a=c\sqrt2$, dengan c merupakan bilangan bulat positif.
\(\begin{aligned} b^2&=(c\sqrt2)^2+2(c\sqrt2)\sqrt2+2\\ n&=2c^2+4c+2\\ n&=2(c+1)^2\leq50\\ c+1&\leq5\\ c&\leq4 \end{aligned}\)
\(\begin{aligned} \dfrac{a^2+b^2}2-ab&=1&{\color{red}\times2}\\[4pt] a^2+b^2-2ab&=2\\ (b-a)^2&=2\\ b-a&=\sqrt2\\ b&=a+\sqrt2\\ b^2&=a^2+2a\sqrt2+2 \end{aligned}\)
Perhatikan bahwa
\(\begin{aligned} b^2&=(c\sqrt2)^2+2(c\sqrt2)\sqrt2+2\\ n&=2c^2+4c+2\\ n&=2(c+1)^2\leq50\\ c+1&\leq5\\ c&\leq4 \end{aligned}\)
Jadi, banyaknya pasangan bilangan bulat positif m dan n dengan m < n ≤ 50 yang memenuhi $$\dfrac{m+n}2-\sqrt{mn}=1$$ adalah 4 pasangan.
JAWAB: D
JAWAB: D
No.
Jika $x^{\frac13}+x^{-\frac13}=2$, maka nilai dari $x^2+\dfrac1{x^2}=$ ....ALTERNATIF PENYELESAIAN
misal $x^{\frac13}=p$,
$p+\dfrac1p=2$
\(\begin{aligned} p^2+\dfrac1{p^2}&=\left(p+\dfrac1p\right)^2-2\\[4pt] &=2^2-2\\ &=2 \end{aligned}\)
\(\begin{aligned} p^4+\dfrac1{p^4}&=\left(p^2+\dfrac1{p^2}\right)^2-2\\[4pt] &=2 \end{aligned}\)
\(\begin{aligned} \left(p^2+\dfrac1{p^2}\right)\left(p^4+\dfrac1{p^4}\right)&=(2)(2)\\[4pt] p^6+\dfrac1{p^6}+p^2+\dfrac1{p^2}&=4\\[4pt] p^6+\dfrac1{p^6}+2&=4\\[4pt] p^6+\dfrac1{p^6}&=4-2\\ x^2+\dfrac1{x^2}&=\color{blue}\boxed{\boxed{\color{black}2}} \end{aligned}\)
$p+\dfrac1p=2$
\(\begin{aligned} p^2+\dfrac1{p^2}&=\left(p+\dfrac1p\right)^2-2\\[4pt] &=2^2-2\\ &=2 \end{aligned}\)
\(\begin{aligned} p^4+\dfrac1{p^4}&=\left(p^2+\dfrac1{p^2}\right)^2-2\\[4pt] &=2 \end{aligned}\)
\(\begin{aligned} \left(p^2+\dfrac1{p^2}\right)\left(p^4+\dfrac1{p^4}\right)&=(2)(2)\\[4pt] p^6+\dfrac1{p^6}+p^2+\dfrac1{p^2}&=4\\[4pt] p^6+\dfrac1{p^6}+2&=4\\[4pt] p^6+\dfrac1{p^6}&=4-2\\ x^2+\dfrac1{x^2}&=\color{blue}\boxed{\boxed{\color{black}2}} \end{aligned}\)
Jadi,
JAWAB:
JAWAB:
No.
Diberikan bilangan real x yang memenuhi $\left(x+\dfrac1x\right)^2=3$. Bentuk sederhana dari- x
- 6
- x + x2
- 0
- 3
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(x+\dfrac1x\right)^2&=3\\[4pt]
x^2+\dfrac1{x^2}+2&=3\\[4pt]
x^2+\dfrac1{x^2}&=1\\[4pt]
x^4+1&=x^2\\
x^4&=x^2-1\\
x^6&=x^4-x^2\\
x^6&=-1
\end{aligned}\)
\(\begin{aligned} x^{2017}+x^{67}+x^{44}+x^{31}+x^{26}+x+6&=\left(x^6\right)^{336}\cdot x+\left(x^6\right)^{11}\cdot x+\left(x^6\right)^7\cdot x^2+\left(x^6\right)^5\cdot x+\left(x^6\right)^4\cdot x^2+x+6\\ &=\left(-1\right)^{336}\cdot x+\left(-1\right)^{11}\cdot x+\left(-1\right)^7\cdot x^2+\left(-1\right)^5\cdot x+\left(-1\right)^4\cdot x^2+x+6\\ &=x-x-x^2-x+x^2+x+6\\ &=\color{blue}\boxed{\boxed{\color{black}6}} \end{aligned}\)
\(\begin{aligned} x^{2017}+x^{67}+x^{44}+x^{31}+x^{26}+x+6&=\left(x^6\right)^{336}\cdot x+\left(x^6\right)^{11}\cdot x+\left(x^6\right)^7\cdot x^2+\left(x^6\right)^5\cdot x+\left(x^6\right)^4\cdot x^2+x+6\\ &=\left(-1\right)^{336}\cdot x+\left(-1\right)^{11}\cdot x+\left(-1\right)^7\cdot x^2+\left(-1\right)^5\cdot x+\left(-1\right)^4\cdot x^2+x+6\\ &=x-x-x^2-x+x^2+x+6\\ &=\color{blue}\boxed{\boxed{\color{black}6}} \end{aligned}\)
Jadi, bentuk sederhana dari x2017 + x67 + x44 + x31 + x26 + x + 6 adalah 6.
JAWAB: B
JAWAB: B
No.
Dalam suatu permainan, jika menang mendapat nilai 1, jika kalah mendapat nilai −1.ALTERNATIF PENYELESAIAN
Misalkan banyaknya menang = x, x ≥ 0
Banyaknya kalah = y, y ≥ 0
sehingga
\(\begin{aligned} x+y&=a=20\\ x-y&=b&\qquad+\\\hline 2x&=20+b\\ b&=2x-20 \end{aligned}\)
Karena0 ≤ x ≤ 20, maka ada 21 kemungkinan nilai x.
Banyaknya kalah = y, y ≥ 0
sehingga
\(\begin{aligned} x+y&=a=20\\ x-y&=b&\qquad+\\\hline 2x&=20+b\\ b&=2x-20 \end{aligned}\)
Karena
Jadi, ada 21 kemungkinan (a, b) pada putaran ke 20.
No.
Diketahui a, b, c adalah bilangan riil dengan $\dfrac{a}{c}=8$. Nilai terkecil dari $\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}$ adalah ....- 1
- 4
- 8
- 16
- 65
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(\dfrac{a}{b}-\dfrac{b}{c}\right)^2&\geq0\\
\dfrac{a^2}{b^2}-2\dfrac{a}{b}\dfrac{b}{c}+\dfrac{b^2}{c^2}&\geq0\\
\dfrac{a^2}{b^2}-2\dfrac{a}{c}+\dfrac{b^2}{c^2}&\geq0\\
\dfrac{a^2}{b^2}-16+\dfrac{b^2}{c^2}&\geq0\\
\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}&\geq16
\end{aligned}\)
Jadi, nilai terkecil dari $\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}$ adalah 16.
JAWAB: D
JAWAB: D
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