Question

[Phạm Tú Uyên]

Bài 1: Thực hiện phép tính

a. \(\frac{11x+10}{3x-3}+\frac{15x+13}{4-4x}\)

b. \(\frac{5x+3}{x^2-3x}+\frac{9-x}{9-3x}\)

c. \(\frac{4xy-1}{5x^2y}-\frac{2xy-1}{5x^2y}\)

d. \(\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}\)

e. \(\frac{x^2-49}{2x+1}.\frac{3}{7-x}\)

f. \(\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{\left(2-3x\right)^3}\)

g. \(\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}\)

h. \(\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}\)

Answer 1

[Akai Haruma]

f)

$\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{(2-3x)^3}$

$=\frac{2x-3x^2}{x^2-1}.\frac{x^4-1}{(2-3x)^3}=\frac{x(2-3x)(x^2-1)(x^2+1)}{(x^2-1)(2-3x)^3}$

$=\frac{x(x^2+1)}{(2-3x)^2}$
g)

$\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15xy^3}$

$=\frac{5xy}{2x-3}.\frac{-4(2x-3)}{15xy^3}=\frac{-4}{3y^2}$

h)

$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$

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

$=\frac{5x}{6(x-1)}$

Answer 2

[Akai Haruma]

d)

$\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}=\frac{x+8}{(x-4)(x+4)}-\frac{2}{x(x+4)}$

$=\frac{x(x+8)}{x(x-4)(x+4)}-\frac{2(x-4)}{x(x+4)(x-4)}$

$=\frac{x^2+8x-2(x-4)}{x(x+4)(x-4)}=\frac{x^2+6x+8}{x(x+4)(x-4)}$

$=\frac{(x+2)(x+4)}{x(x+4)(x-4)}=\frac{x+2}{x(x-4)}$
e)

$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$

$=\frac{-3(x+7)}{2x+1}$