[tex]a)\frac{11x+3}{3x-3}+\frac{15x+17}{4-4x}[/tex]
b) [tex]\frac{1}{x^2+x+1}+\frac{1}{x^2-x}+\frac{2x}{1-x^3}[/tex]
c)[tex]\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}[/tex]
d)[tex]\frac{3x^2+5x+1}{x^3-1}-\frac{1-x}{x^2+x+1}-\frac{3}{x-1}[/tex]
đề như vậy nè bạn
nhờ làm lại giúp mình với
a) ĐK: $x\ne 1$.
$\dfrac{11x+3}{3x-3}+\dfrac{15x+17}{4-4x}
\\=\dfrac{11x+3}{3(x-1)}-\dfrac{15x+17}{4(x-1)}
\\=\dfrac{4(11x+3)-3(15x+17)}{12(x-1)}
\\=\dfrac{44x+12-45x-51}{12(x-1)}
\\=\dfrac{-x-39}{12(x-1)}
\\=\dfrac{x+39}{12-12x}$
b) ĐK: $x\ne 0;x\ne 1$.
$\dfrac{1}{x^2+x+1}+\dfrac{1}{x^2-x}+\dfrac{2x}{1-x^3}
\\=\dfrac{1}{x^2+x+1}+\dfrac{1}{x(x-1)}-\dfrac{2x}{(x-1)(x^2+x+1)}
\\=\dfrac{x(x-1)+x^2+x+1-2x^2}{x(x-1)(x^2+x+1)}
\\=\dfrac{x^2-x+x^2+x+1-2x^2}{x(x^3-1)}
\\=\dfrac{1}{x^4-x}$
c) ĐK: $x\ne 0;x\ne \pm 3$.
$\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}
\\=\dfrac{x+9}{(x-3)(x+3)}-\dfrac{3}{x(x+3)}
\\=\dfrac{x(x+9)-3(x-3)}{x(x-3)(x+3)}
\\=\dfrac{x^2+9x-3x+9}{x(x-3)(x+3)}
\\=\dfrac{x^2+6x+9}{x(x-3)(x+3)}
\\=\dfrac{(x+3)^2}{(x^2-3x)(x+3)}
\\=\dfrac{x+3}{x^2-3x}$
d) ĐK: $x\ne 1$.
$\dfrac{3x^2+5x+1}{x^3-1}-\dfrac{1-x}{x^2+x+1}-\dfrac{3}{x-1}
\\=\dfrac{3x^2+5x+1}{(x-1)(x^2+x+1)}+\dfrac{x-1}{x^2+x+1}-\dfrac{3}{x-1}
\\=\dfrac{3x^2+5x+1+(x-1)^2-3(x^2+x+1)}{(x-1)(x^2+x+1)}
\\=\dfrac{3x^2+5x+1+x^2-2x+1-3x^2-3x-3}{(x-1)(x^2+x+1)}
\\=\dfrac{x^2-1}{(x-1)(x^2+x+1)}
\\=\dfrac{(x-1)(x+1)}{(x-1)(x^2+x+1)}
\\=\dfrac{x+1}{x^2+x+1}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
