Question

Quy đồng mẫu thức các phân thức :

a.\(\dfrac{x+3}{x+1}\);\(\dfrac{2x-1}{x-1}\);\(\dfrac{2x^2-3x}{x^2-1}\)

b.\(\dfrac{2\left(6x-5\right)}{x^2-10x+25}\);\(\dfrac{3x}{x-5}\);\(\dfrac{x-2}{x+5}\)

c.\(\dfrac{3x}{x+3}\);\(\dfrac{x-3}{x-1}\);\(\dfrac{5x-13}{\left(1-x\right)\left(x+3\right)}\)

d.\(\dfrac{1}{2x+2}\);\(\dfrac{1}{2x-2}\);\(\dfrac{1}{1-x^2}\)

e.\(\dfrac{1}{3x+3y}\);\(\dfrac{2x}{x^2-y^2}\)và\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}\)

f.\(\dfrac{1}{x+2}\);\(\dfrac{x+1}{x^2-4x+4}\)và\(\dfrac{5}{2-x}\)

Answer 1

a) MTC: \(\left(x-1\right)\left(x+1\right)\)

\(\dfrac{x+3}{x+1}=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)}\)

\(\dfrac{2x-1}{x-1}=\dfrac{\left(2x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{2x^2-3x}{x^2-1}=\dfrac{x\left(2x-3\right)}{\left(x-1\right)\left(x+1\right)}\)

--------

b) MTC: \(\left(x+5\right)\left(x-5\right)^2\)

\(\dfrac{2\left(6x-5\right)}{\left(x^2-10x+25\right)}=\dfrac{2\left(6x-5\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)^2}\)

\(\dfrac{3x}{x-5}=\dfrac{3x\left(x-5\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)^2}\)

\(\dfrac{x-2}{x+5}=\dfrac{\left(x-2\right)\left(x-5\right)^2}{\left(x+5\right)\left(x-5\right)^2}\)

--------

c) MTC: \(\left(x-1\right)\left(x+3\right)\)

\(\dfrac{3x}{x+3}=\dfrac{3x\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)

\(\dfrac{x-3}{x-1}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)

\(\dfrac{5x-13}{\left(1-x\right)\left(x+3\right)}=\dfrac{13-5x}{\left(x-1\right)\left(x+3\right)}\)

--------

d) MTC: \(2\left(x-1\right)\left(x+1\right)\)

\(\dfrac{1}{2x+2}=\dfrac{1}{2\left(x+1\right)}=\dfrac{1\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{1}{2x-2}=\dfrac{1}{2\left(x-1\right)}=\dfrac{1.\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{2\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{1}{1-x^2}=-\dfrac{1}{x^2-1}=-\dfrac{1.2}{2\left(x-1\right)\left(x+1\right)}=-\dfrac{2}{2\left(x-1\right)\left(x+1\right)}\)

--------

e) MTC: \(3\left(x+y\right)\left(x-y\right)^2\)

\(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{1.\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)

\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{2x.3}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{6x}{3\left(x+y\right)\left(x-y\right)^2}\)

\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{3\left(x+y\right)\left(x^2-xy+y^2\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)

--------

f) MTC: \(\left(x+2\right)\left(x-2\right)^2\)

\(\dfrac{1}{x+2}=\dfrac{1.\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)

\(\dfrac{x+1}{x^2-4x+4}=\dfrac{x+1}{\left(x-2\right)^2}=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)^2}\)

\(\dfrac{5}{2-x}=-\dfrac{5}{x-2}=-\dfrac{5\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)^2}\)