Teka-teki Persamaan Tiga Variabel

Soal:

Diketahui

\displaystyle abc=10

\displaystyle a+b+c=-7

\displaystyle ab+ac+bc=2

Ditanyakan

\displaystyle \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\ldots

Jawaban:

Untuk bisa menjawab soal ini, Anda harus mau mengotak-atik bentuk pecahan dengan variabel. Mulai saja dengan data yang tersedia.

Kita mulai dengan menyamakan penyebut pada penjumlahan \displaystyle \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}.

\displaystyle \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\left( \frac{ab}{c}\times \frac{ab}{ab} \right)+\left( \frac{ac}{b}\times \frac{ac}{ac} \right)+\left( \frac{bc}{a}\times \frac{bc}{bc} \right)

\displaystyle \Leftrightarrow \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\frac{{{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}}{abc}

Persamaan \displaystyle {{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}} bisa kita uraikan menjadi

\displaystyle {{\left( ab+ac+bc \right)}^{2}}=\left( ab+ac+bc \right)\left( ab+ac+bc \right)

\displaystyle \Leftrightarrow {{\left( ab+ac+bc \right)}^{2}}=\left( ab\left( ab+ac+bc \right)+ac\left( ab+ac+bc \right)+bc\left( ab+ac+bc \right) \right)

\displaystyle \Leftrightarrow {{\left( ab+ac+bc \right)}^{2}}=\left( {{a}^{2}}{{b}^{2}}+{{a}^{2}}bc+a{{b}^{2}}c \right)+\left( {{a}^{2}}bc+{{a}^{2}}{{c}^{2}}+ab{{c}^{2}} \right)+\left( a{{b}^{2}}c+ab{{c}^{2}}+{{b}^{2}}{{c}^{2}} \right)

\displaystyle \Leftrightarrow {{\left( ab+ac+bc \right)}^{2}}=\left( {{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}} \right)+\left( {{a}^{2}}bc+a{{b}^{2}}c+{{a}^{2}}bc+ab{{c}^{2}}+a{{b}^{2}}c+ab{{c}^{2}} \right)

\displaystyle \Leftrightarrow {{\left( ab+ac+bc \right)}^{2}}=\left( {{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}} \right)+\left( {{a}^{2}}bc+a{{b}^{2}}c+{{a}^{2}}bc+ab{{c}^{2}}+a{{b}^{2}}c+ab{{c}^{2}} \right)

\displaystyle \Leftrightarrow {{\left( ab+ac+bc \right)}^{2}}=\left( {{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}} \right)+\left( 2{{a}^{2}}bc+2a{{b}^{2}}c+2ab{{c}^{2}} \right)

\displaystyle \Leftrightarrow {{\left( ab+ac+bc \right)}^{2}}=\left( {{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}} \right)+2\left( {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}} \right)

\displaystyle \Leftrightarrow {{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}={{\left( ab+ac+bc \right)}^{2}}-2\left( {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}} \right) ………..(1)

Persamaan \displaystyle {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}} kita ubah ke dalam bentuk pecahan agar nilainya bisa dihitung.

\displaystyle {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}}=\left( {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}} \right)\times \frac{abc}{abc}

\displaystyle \Leftrightarrow {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}}=\left( \frac{{{a}^{2}}bc}{abc}+\frac{a{{b}^{2}}c}{abc}+\frac{ab{{c}^{2}}}{abc} \right)\times abc

\displaystyle \Leftrightarrow {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}}=\left( \frac{a\cdot abc}{abc}+\frac{b\cdot abc}{abc}+\frac{c\cdot abc}{abc} \right)\times abc

\displaystyle \Leftrightarrow {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}}=\left( a+b+c \right)\times abc ………..(2)

Selanjutnya, kita bisa  memasukkan (2) ke (1).

\displaystyle {{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}={{\left( ab+ac+bc \right)}^{2}}-2\left( \left( a+b+c \right)\times abc \right)

Hitung soal yang diberikan menggunakan informasi yang sudah diketahui.

\displaystyle \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\frac{{{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}}{abc}

\displaystyle \Leftrightarrow \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\frac{{{\left( ab+ac+bc \right)}^{2}}-2\left( \left( a+b+c \right)\times abc \right)}{abc}

\displaystyle \Leftrightarrow \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\frac{{{\left( 2 \right)}^{2}}-2\left( \left( -7 \right)\times 10 \right)}{10}

\displaystyle \Leftrightarrow \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\frac{4-2\left( -70 \right)}{10}

\displaystyle \Leftrightarrow \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\frac{144}{10}

\displaystyle \Leftrightarrow \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=\frac{72}{5}

 

 

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