Tolong bantu soal matematika yg di gambar ya !.... Question from @cantiqahhhjjjn

September 2018 1 16 Report

Tolong bantu soal matematika yg di gambar ya !

Matrixisme A) $\frac{4}{x+3}$ - $\frac{5}{x-1}$ = $\frac{4(x-1)}{(x+3)(x-1)}$ - $\frac{5(x+3)}{(x-1)(x+3)}$ = $\frac{4(x-1)-(5(x+3))}{(x+3)(x-1)}$ = $\frac{4x-4-5x-15}{x^{2}-x+3x-3}$ = $\frac{-x-19}{x^{2}+2x-3}$

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

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

d) $\frac{x^{2}+4x-12}{2x^{2}+9x-18}$ = $\frac{(x-2)(x+6)}{(2x-3)(x+6)}$ = $\frac{x-2}{2x-3}$

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

g) $\frac{12}{x^{2}-9}$ : $\frac{3}{x+3}$ = $\frac{12}{(x-3)(x+3)}$ x $\frac{x+3}{3}$ = $\frac{4}{x-3}$

h) $\frac{\frac{1}{x+y}-\frac{2}{x-y}}{\frac{3}{x-y}+\frac{4}{x+y}}$ = $\frac{\frac{1(x-y)}{(x+y)(x-y)}-\frac{2(x+y)}{(x-y)(x+y)}}{\frac{3(x+y)}{(x-y)(x+y)}+\frac{4(x-y)}{(x+y)(x-y)}}$ = $\frac{\frac{x-y-(2x+2y)}{(x-y)(x+y)}}{\frac{3x+3y+4x-4y}{(x+y)(x-y)}}$ = $\frac{x-y-2x-2y}{(x-y)(x+y)}$ : $\frac{3x+3y+4x-4y}{(x+y)(x-y)}$ = $\frac{x-y-2x-2y}{(x-y)(x+y)}$ x $\frac{(x+y)(x-y)}{3x+3y+4x-4y}$ = $\frac{x-y-2x-2y}{3x+3y+4x-4y}$ = $\frac{-x-3y}{7x-y}$

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Matrixisme sama2,....