HOTS Zone: Sistem Persamaan Aljabar [3]
Table of Contents
Tipe:
No.
Diketahui sistem persamaan \begin{align}\dfrac1a+\dfrac1{b+c}=2\\[4pt]\dfrac1b+\dfrac1{a+c}=3\\[4pt]\dfrac1c+\dfrac1{a+b}=4\end{align} JikaALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac1a+\dfrac1{b+c}&=2\\[4pt]
\dfrac{a+b+c}{ab+ac}&=2\\[4pt]
a+b+c&=2(ab+ac)\\
6(a+b+c)&=12(ab+ac)&\qquad{\color{red}(4)}
\end{aligned}\)
Dengan cara yang sama, dari persamaan (2) dan (3) didapat:
\(\begin{aligned} 4(a+b+c)&=12(ab+bc)&\qquad{\color{red}(5)}\\ 3(a+b+c)&=12(ac+bc)&\qquad{\color{red}(6)} \end{aligned}\)
Jumlahkan persamaan (4), (5), dan (6) didapat:
\(\begin{aligned} 13(a+b+c)&=12(2ab+2bc+2ac)\\ 13(a+b+c)&=24(ab+bc+ac)\\ ab+bc+ac&=\dfrac{13}{24}(a+b+c) \end{aligned}\)
\(\begin{aligned} a^2+b^2+c^2&=10\\ (a+b+c)^2-2(ab+bc+ac)&=10&{\color{red}\times12}\\ 12(a+b+c)^2-24(ab+bc+ac)&=120\\ 12(a+b+c)^2-13(a+b+c)-120&=0\\ 12x^2-13x-120&=0\\ (4x-15)(3x+8)&=0 \end{aligned}\)
Dengan cara yang sama, dari persamaan (2) dan (3) didapat:
\(\begin{aligned} 4(a+b+c)&=12(ab+bc)&\qquad{\color{red}(5)}\\ 3(a+b+c)&=12(ac+bc)&\qquad{\color{red}(6)} \end{aligned}\)
Jumlahkan persamaan (4), (5), dan (6) didapat:
\(\begin{aligned} 13(a+b+c)&=12(2ab+2bc+2ac)\\ 13(a+b+c)&=24(ab+bc+ac)\\ ab+bc+ac&=\dfrac{13}{24}(a+b+c) \end{aligned}\)
\(\begin{aligned} a^2+b^2+c^2&=10\\ (a+b+c)^2-2(ab+bc+ac)&=10&{\color{red}\times12}\\ 12(a+b+c)^2-24(ab+bc+ac)&=120\\ 12(a+b+c)^2-13(a+b+c)-120&=0\\ 12x^2-13x-120&=0\\ (4x-15)(3x+8)&=0 \end{aligned}\)
- 4x − 15 = 0
\(\begin{aligned} x&=\dfrac{15}4\\ a+b+c&=\color{blue}\boxed{\boxed{\color{black}\dfrac{15}4}} \end{aligned}\) - 3x + 8 = 0
\(\begin{aligned} x&=-\dfrac83\\ a+b+c&=\color{blue}\boxed{\boxed{\color{black}-\dfrac83}} \end{aligned}\)
Jadi, $a+b+c=\dfrac{15}4$ atau $-\dfrac83$.
No.
Bilangan real t sehingga terdapat dengan tunggal tripel bilangan realALTERNATIF PENYELESAIAN
\(\begin{aligned}
x^2+2y^2&=3z\\
3x+3y+3z&=3t&\ +\\\hline
x^2+3x+2y^2+3y&=3t\\
x^2+3x+\dfrac94+2y^2+3y+\dfrac9{8}&=3t+\dfrac94+\dfrac9{8}\\[4pt]
\left(x+\dfrac32\right)^2+2\left(y+\dfrac34\right)^2&=3t+\dfrac{27}{8}
\end{aligned}\)
Agar(x, y, z) tunggal, maka
\(\begin{aligned} 3t+\dfrac{27}{8}&=0\\[4pt] t&=\color{blue}\boxed{\boxed{\color{black}-\dfrac98}} \end{aligned}\)
Agar
\(\begin{aligned} 3t+\dfrac{27}{8}&=0\\[4pt] t&=\color{blue}\boxed{\boxed{\color{black}-\dfrac98}} \end{aligned}\)
Jadi, bilangan real t sehingga terdapat dengan tunggal tripel bilangan real (x, y, z) yang memenuhi x2 + 2y2 = 3z dan x + y + z = t adalah $-\dfrac98$.
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