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a)11x+13/3-3+(15x+17/4-4x)
b)1/x^2+x+11+(1/x^2-x+2x/1-x^3)
c)x+9/x^2-9-(3/x^2+3x)
d)3x^2+5x+1/x^3-1-(1-x/x^2+x+1)-3/x-1

 

[Toshiro Koyoshi]

 

[Thái Sơn Long]

View attachment 24205

$$a)\frac{11x+3}{3x-3}+\frac{15x+17}{4-4x}$$
b) $$\frac{1}{x^2+x+1}+\frac{1}{x^2-x}+\frac{2x}{1-x^3}$$
c)$$\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}$$
d)$$\frac{3x^2+5x+1}{x^3-1}-\frac{1-x}{x^2+x+1}-\frac{3}{x-1}$$
đề như vậy nè bạn
nhờ làm lại giúp mình với

 

[Thái Sơn Long]

@Nữ Thần Mặt Trăng
giúp e vs c ơi

 

[Nữ Thần Mặt Trăng]

$$a)\frac{11x+3}{3x-3}+\frac{15x+17}{4-4x}$$
b) $$\frac{1}{x^2+x+1}+\frac{1}{x^2-x}+\frac{2x}{1-x^3}$$
c)$$\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}$$
d)$$\frac{3x^2+5x+1}{x^3-1}-\frac{1-x}{x^2+x+1}-\frac{3}{x-1}$$
đề như vậy nè bạn
nhờ làm lại giúp mình với

@Nữ Thần Mặt Trăng
giúp e vs c ơi

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: $x\ne 1$.
$\dfrac{3x^2+5x+1}{x^3-1}-\dfrac{1-x}{x^2+x+1}-\dfrac{3}{x-1}
\\=\dfrac{3x^2+5x+1}{(x-1)(x^2+x+1)}+\dfrac{x-1}{x^2+x+1}-\dfrac{3}{x-1}
\\=\dfrac{3x^2+5x+1+(x-1)^2-3(x^2+x+1)}{(x-1)(x^2+x+1)}
\\=\dfrac{3x^2+5x+1+x^2-2x+1-3x^2-3x-3}{(x-1)(x^2+x+1)}
\\=\dfrac{x^2-1}{(x-1)(x^2+x+1)}
\\=\dfrac{(x-1)(x+1)}{(x-1)(x^2+x+1)}
\\=\dfrac{x+1}{x^2+x+1}$

 

[Thái Sơn Long]

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: $x\ne 1$.
$\dfrac{3x^2+5x+1}{x^3-1}-\dfrac{1-x}{x^2+x+1}-\dfrac{3}{x-1}
\\=\dfrac{3x^2+5x+1}{(x-1)(x^2+x+1)}+\dfrac{x-1}{x^2+x+1}-\dfrac{3}{x-1}
\\=\dfrac{3x^2+5x+1+(x-1)^2-3(x^2+x+1)}{(x-1)(x^2+x+1)}
\\=\dfrac{3x^2+5x+1+x^2-2x+1-3x^2-3x-3}{(x-1)(x^2+x+1)}
\\=\dfrac{x^2-1}{(x-1)(x^2+x+1)}
\\=\dfrac{(x-1)(x+1)}{(x-1)(x^2+x+1)}
\\=\dfrac{x+1}{x^2+x+1}$

câu 1 em ghi nhầm đề
là 11x+13 nhé c

 

[Nữ Thần Mặt Trăng]

câu 1 em ghi nhầm đề
là 11x+13 nhé c

a) ĐK: $x\ne 1$.
$\dfrac{11x+13}{3x-3}+\dfrac{15x+17}{4-4x}
\\=\dfrac{11x+13}{3(x-1)}-\dfrac{15x+17}{4(x-1)}
\\=\dfrac{4(11x+13)-3(15x+17)}{12(x-1)}
\\=\dfrac{44x+52-45x-51}{12(x-1)}
\\=\dfrac{1-x}{12(x-1)}
\\=\dfrac{-1}{12}$