HOTS Zone : Aljabar [2]
Tipe:
No.
Diketahuimaka nilai xy =
- 4066272
- 4068289
- 5750577,011
- 5756281,95
- 8132544
ALTERNATIF PENYELESAIAN
Misal \(a=\sqrt{(x+2016)(x-2016)}\)
\(\begin{aligned} a^2&=x^2-2016^2\\ x^2&=a^2+2016^2 \end{aligned}\)
Misal \(b=\sqrt{(y+2017)(y-2017)}\)
\(\begin{aligned} b^2&=y^2-2017^2\\ y^2&=b^2+2017^2 \end{aligned}\)
\(\begin{aligned} 2016a+2017b&=\dfrac12\left(a^2+2016^2+b^2+2017^2\right)\\ 2\cdot2016a+2\cdot2017b&=a^2+2016^2+b^2+2017^2\\ a^2-2\cdot2016a+2016^2+b^2-2\cdot2017b+2017^2&=0\\ (a-2016)^2+(b-2017)^2&=0 \end{aligned}\)
didapata= 2016 dan b= 2017
\(\begin{aligned} x^2&=2016^2+2016^2\\ &=2\cdot2016^2\\ x&=2016\sqrt2 \end{aligned}\)
\(\begin{aligned} y^2&=2017^2+2017^2\\ &=2\cdot2017^2\\ x&=2017\sqrt2 \end{aligned}\)
\(\begin{aligned} xy&=2016\sqrt2\cdot2017\sqrt2\\ &=\boxed{\boxed{8132544}} \end{aligned}\)
\(\begin{aligned} a^2&=x^2-2016^2\\ x^2&=a^2+2016^2 \end{aligned}\)
Misal \(b=\sqrt{(y+2017)(y-2017)}\)
\(\begin{aligned} b^2&=y^2-2017^2\\ y^2&=b^2+2017^2 \end{aligned}\)
\(\begin{aligned} 2016a+2017b&=\dfrac12\left(a^2+2016^2+b^2+2017^2\right)\\ 2\cdot2016a+2\cdot2017b&=a^2+2016^2+b^2+2017^2\\ a^2-2\cdot2016a+2016^2+b^2-2\cdot2017b+2017^2&=0\\ (a-2016)^2+(b-2017)^2&=0 \end{aligned}\)
didapat
\(\begin{aligned} x^2&=2016^2+2016^2\\ &=2\cdot2016^2\\ x&=2016\sqrt2 \end{aligned}\)
\(\begin{aligned} y^2&=2017^2+2017^2\\ &=2\cdot2017^2\\ x&=2017\sqrt2 \end{aligned}\)
\(\begin{aligned} xy&=2016\sqrt2\cdot2017\sqrt2\\ &=\boxed{\boxed{8132544}} \end{aligned}\)
Jadi, xy = 8132544.
JAWAB: E
JAWAB: E
No.
Jika \(n-\dfrac1n=x\), maka berapakah \(n^2+\dfrac1{n^2}\) jika dinyatakan dalam x?- x2 + 1
- x + 1
- x3 + 1
- x2 + 2
- \(\sqrt{x}+\sqrt1\)
ALTERNATIF PENYELESAIAN
\(\eqalign{
n-\dfrac1n&=x\\
\left(n-\dfrac1n\right)^2&=x^2\\
n^2-2(n)\left(\dfrac1n\right)+\left(\dfrac1n\right)^2&=x^2\\
n^2-2+\dfrac1{n^2}&=x^2\\
n^2+\dfrac1{n^2}&=\boxed{\boxed{x^2+2}}
}\)
Jadi, \(n^2+\dfrac1{n^2}=x^2+2\).
JAWAB: D
JAWAB: D
No.
Jika \(\dfrac{3a+4b}{2a-2b}=5\) maka tentukan nilai dari \(\dfrac{a^2+2b^2}{ab}\)ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{3a+4b}{2a-2b}&=5\\
3a+4b&=10a-10b\\
7a&=14b\\
\dfrac{a}b&=2
\end{aligned}\)
\(\begin{aligned}
\dfrac{a^2+2b^2}{ab}&=\dfrac{a}b+2\dfrac{b}a\\
&=2+2\left(\dfrac12\right)\\
&=\boxed{\boxed{3}}
\end{aligned}\)
Jadi, \(\dfrac{a^2+2b^2}{ab}=3\).
No.
Nilai dari \(\dfrac{(1945+2011)^2+(2011-1945)^2}{1945^2+2011^2}\) adalah ....ALTERNATIF PENYELESAIAN
Misal a = 1945 dan b = 2011
\(\begin{aligned} \dfrac{(1945+2011)^2+(2011-1945)^2}{1945^2+2011^2}&=\dfrac{(a+b)^2+(b-a)^2}{a^2+b^2}\\ &=\dfrac{a^2+2ab+b^2+b^2-2ab+a^2}{a^2+b^2}\\ &=\dfrac{2\left(a^2+b^2\right)}{a^2+b^2}\\ &=\boxed{\boxed{2}} \end{aligned}\)
\(\begin{aligned} \dfrac{(1945+2011)^2+(2011-1945)^2}{1945^2+2011^2}&=\dfrac{(a+b)^2+(b-a)^2}{a^2+b^2}\\ &=\dfrac{a^2+2ab+b^2+b^2-2ab+a^2}{a^2+b^2}\\ &=\dfrac{2\left(a^2+b^2\right)}{a^2+b^2}\\ &=\boxed{\boxed{2}} \end{aligned}\)
Jadi, \(\dfrac{(1945+2011)^2+(2011-1945)^2}{1945^2+2011^2}=2.\)
No.
Bilangan-bilangan real a, b, dan c memenuhi sistem persamaan- 4
- 5
- 6
- 7
- 8
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
ab&\leq\dfrac{(a+b)^2}4\\
&\leq\dfrac{8^2}4\\
&\leq16
\end{aligned}\)
\(\begin{aligned} c^2+16&\geq16\\ ab&\geq16\\ \end{aligned}\)
Didapat ab = 16 dan c = 0
a + b + c = 8 + 0 = 8
\(\begin{aligned} c^2+16&\geq16\\ ab&\geq16\\ \end{aligned}\)
Didapat ab = 16 dan c = 0
a + b + c = 8 + 0 = 8
Jadi, a + b + c = 8.
JAWAB: E
JAWAB: E
No.
Jika \(x+\dfrac1x=4\) maka nilaiALTERNATIF PEMBAHASAN
\(\eqalign{
\left(x+\dfrac1x\right)^2&=4^2\\
x^2+2+\dfrac1{x^2}&=16\\
x^2+\dfrac1{x^2}&=14
}\)
\(\eqalign{
\left(x^2+\dfrac1{x^2}\right)\left(x+\dfrac1x\right)&=(14)(4)\\
x^3+x+\dfrac1x+\dfrac1{x^3}&=56\\
x^3+4+\dfrac1{x^3}&=56\\
x^3+\dfrac1{x^3}&=\boxed{\boxed{52}}
}\)
Jadi, x3 + x−3 = 52 .
No.
Koefisien x2 pada ekspansi- −2
- −1
- 2
- 7
- 9
ALTERNATIF PENYELESAIAN
Kita ambil satu suku dari setiap kurung sedemikian sehingga hasil kalinya mengandung variabel x2.
\(\begin{aligned} \left(x^2\right)(-1)+(3x)(3x)+(-1)\left(x^2\right)&=-x^2+9x^2-x^2\\ &=7x^2 \end{aligned}\)
Jadi, koefisien x2 pada ekspansi (x2 + 3x − 1)2 adalah 7.
JAWAB: D
JAWAB: D
No.
Diketahui x y memenuhi \[\dfrac{(x-2023)(y-2024)}{(x-2023)^2(y-2024)^2}=-\dfrac12\] Nilai yang mungkin bagiALTERNATIF PENYELESAIAN
Misal a = x − 2023 , dan b = y − 2023
\(\begin{aligned} \dfrac{ab}{a^2+b^2}&=-\dfrac12\\[3.8pt] 2ab&=-a^2-b^2\\ a^2+2ab+b^2&=0\\ (a+b)^2&=0\\ a+b&=0\\ x-2023+y-2024&=0\\ x+y&=4047 \end{aligned}\)
\(\begin{aligned} \dfrac{ab}{a^2+b^2}&=-\dfrac12\\[3.8pt] 2ab&=-a^2-b^2\\ a^2+2ab+b^2&=0\\ (a+b)^2&=0\\ a+b&=0\\ x-2023+y-2024&=0\\ x+y&=4047 \end{aligned}\)
Jadi, nilai yang mungkin bagi x + y adalah 4047.
No.
Jika x dan y bilangan real yang memenuhi \(\dfrac{x+40}y+\dfrac{569}{xy}=\dfrac{26-y}x\), tentukan nilai dari xy.- −260
- −270
- −280
- −300
- −400
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{x+40}y+\dfrac{569}{xy}&=\dfrac{26-y}x&{\color{red}\times xy}\\[3.8pt]
x^2+40x+569&=26y-y^2\\
x^2+40x+400+y^2-26y+169&=0\\
(x+20)^2+(y-13)^2&=0
\end{aligned}\)
x = −20 dan y = 13
xy = −20×13 = −260
x = −20 dan y = 13
xy = −20×13 = −260
Jadi, xy = −260.
JAWAB: A
JAWAB: A
No.
Jika e, b, c adalah bilangan real yang memenuhiALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}-\dfrac1a-\dfrac1b-\dfrac1c&=\dfrac{a^2+b^2+c^2-ab-bc-ac}{abc}\\[3.8pt]
&=\dfrac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{2abc}\\[3.8pt]
&=\dfrac{(a-b)^2+(b-c)^2+(c-a)^2}{2\cdot60}\\[3.8pt]
&=\dfrac{1^2+1^2+2^2}{120}\\
&=\boxed{\boxed{\dfrac1{20}}}
\end{aligned}\)
Jadi,
JAWAB:
JAWAB:
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