Matematika

Pertanyaan

tolong bantu soal matematika yg di gambar ya !
tolong bantu soal matematika yg di gambar ya !

1 Jawaban

  • a) [tex]\frac{4}{x+3}[/tex] - [tex]\frac{5}{x-1}[/tex] = [tex]\frac{4(x-1)}{(x+3)(x-1)}[/tex] - [tex]\frac{5(x+3)}{(x-1)(x+3)}[/tex] = [tex]\frac{4(x-1)-(5(x+3))}{(x+3)(x-1)}[/tex] = [tex]\frac{4x-4-5x-15}{x^{2}-x+3x-3}[/tex] = [tex]\frac{-x-19}{x^{2}+2x-3}[/tex]

    b) [tex]\frac{3}{2x+5}[/tex] + [tex]\frac{2}{6x^{2}+7x-20}[/tex] = [tex]\frac{3}{2x+5}[/tex] + [tex]\frac{2}{(2x+5)(3x-4)}[/tex] = [tex]\frac{3(3x-4)}{(2x+5)(3x-4)}[/tex] + [tex]\frac{2}{(2x+5)(3x-4)}[/tex] = [tex]\frac{3(3x-4)+2}{(2x+5)(3x-4)}[/tex] = [tex]\frac{9x-12+2}{6x^{2}+7x-20}[/tex] = [tex]\frac{9x-10}{6x^{2}+7x-20}[/tex]

    c) [tex]\frac{2a}{3x}[/tex] x [tex]\frac{2x-6xy}{12a}[/tex] = [tex]\frac{2x-6xy}{3x(6)}[/tex] = [tex]\frac{2x(1-3y)}{18x}[/tex] = [tex]\frac{1-3y}{9}[/tex] = [tex]\frac{1}{9}[/tex] - [tex]\frac{y}{3}[/tex]

    d) [tex]\frac{x^{2}+4x-12}{2x^{2}+9x-18}[/tex] = [tex]\frac{(x-2)(x+6)}{(2x-3)(x+6)}[/tex] = [tex]\frac{x-2}{2x-3}[/tex]

    f) [tex]\frac{-6x^{2}+22x-20}{9x^{2}-25}[/tex] = [tex]\frac{(-2x+4)(3x-5)}{(3x-5)(3x+5)}[/tex] = [tex]\frac{-2x+4}{3x+5}[/tex] = [tex]\frac{4-2x}{3x+5}[/tex]

    g) [tex]\frac{12}{x^{2}-9}[/tex] : [tex]\frac{3}{x+3}[/tex] = [tex]\frac{12}{(x-3)(x+3)}[/tex] x [tex]\frac{x+3}{3}[/tex] = [tex]\frac{4}{x-3}[/tex]

    h) [tex]\frac{\frac{1}{x+y}-\frac{2}{x-y}}{\frac{3}{x-y}+\frac{4}{x+y}}[/tex] = [tex]\frac{\frac{1(x-y)}{(x+y)(x-y)}-\frac{2(x+y)}{(x-y)(x+y)}}{\frac{3(x+y)}{(x-y)(x+y)}+\frac{4(x-y)}{(x+y)(x-y)}}[/tex] = [tex]\frac{\frac{x-y-(2x+2y)}{(x-y)(x+y)}}{\frac{3x+3y+4x-4y}{(x+y)(x-y)}}[/tex] = [tex]\frac{x-y-2x-2y}{(x-y)(x+y)}[/tex] : [tex]\frac{3x+3y+4x-4y}{(x+y)(x-y)}[/tex] = [tex]\frac{x-y-2x-2y}{(x-y)(x+y)}[/tex] x [tex]\frac{(x+y)(x-y)}{3x+3y+4x-4y}[/tex] = [tex]\frac{x-y-2x-2y}{3x+3y+4x-4y}[/tex] = [tex]\frac{-x-3y}{7x-y}[/tex]