Question

a)\(\frac{x+1}{2x+6}\)+\(\frac{2x+3}{x^2+3x}\)

b)\(\frac{3}{2x+6}\)- \(\frac{x-6}{2x^2+6x}\)

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Answer 1

a) Ta có : 2x + 6 = 2( x+ 3 )

x² + 3x = x ( x + 3 )

Nên \(\Rightarrow\)MTC = 2x ( x + 3 )

\(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}=\frac{\left(x+1\right)x+2\left(2x+3\right)}{2x\left(x+3\right)}=\frac{x^2+x+4x+6}{2x\left(x+3\right)}=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}=\frac{x\left(x+2\right)+3\left(x+2\right)}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)

b) Ta có : 2x + 6 = 2( x + 3 )

2x² + 6x = 2x( x + 3 )

Nên : MTC = 2x ( x + 3 )

Do đó : \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{3x-x+6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)

Answer 2

a) \(\frac{x+1}{2x+6}\)+ \(\frac{2x+3}{x^2+3x}\)

= \(\frac{x+1}{2\left(x+3\right)}\)+ \(\frac{2x+3}{x\left(x+3\right)}\)

= \(\frac{\left(x+1\right)x}{2x\left(x+3\right)}\) + \(\frac{\left(2x+3\right).2}{2x\left(x+3\right)}\)

= \(\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)= \(\frac{x^2+5x+6}{2x\left(x+3\right)}\)= \(\frac{x^2+2x+3x+6}{2x\left(x+3\right)}\)

= \(\frac{x\left(x+2\right)+3\left(x+2\right)}{2x\left(x+3\right)}\) = \(\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3 \right)}\) = \(\frac{x+2}{2x}\)

Answer 3

b) \(\frac{3}{2x+6}\) - \(\frac{x-6}{2x^2+6x}\)

= \(\frac{3}{2\left(x+3\right)}\)- \(\frac{x-6}{2x\left(x+3\right)}\)

= \(\frac{3x}{2x\left(x+3\right)}\) - \(\frac{x-6}{2x\left(x+3\right)}\)

= \(\frac{3x-x+6}{2x\left(x+3\right)}\) = \(\frac{2x+6}{2x\left(x+3\right)}\) = \(\frac{2\left(x+3\right)}{2x\left(x+3\right)}\) = \(\frac{2}{2x}\)