tentukan nilai m pada bersamaan bentuk aljabar (2x+3y)(kx+ly)=mx2 +23xy+12y2

dikaliin biasa dulu
[tex](2x+3y)(kx+ly)=(m {x}^{2} +23xy+12 {y}^{2} ) \\ (2x.kx+2x.ly+3y.kx + 3y.ly) =(m {x}^{2} +23xy+12 {y}^{2}) \\ (2k {x}^{2} + 2lxy+3kxy+3l {y}^{2} = (m {x}^{2} +23xy+12 {y}^{2}) \\ (2k {x}^{2} +(2l+3k)xy+3l {y}^{2} = (m {x}^{2} +23xy+12 {y}^{2})
[/tex]


nah ada pola kesamaan, berarti kan
[tex]2k {x}^{2} = m {x}^{2} \\
2k = m[/tex]

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[tex](2l+3k).xy =23.xy \\
2l+2k = 23[/tex]

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[tex]3ly^2 = 12y^2 \\ 3l = 12 \\ l= 4[/tex]


substitusi
[tex]2(4) +3k= 23 \\
3k= 23-8\\
3k = 15 \\
k = 5

\\ \\ m = 2(5) \\
m = 10[/tex]


semoga membantu:)