Toán Lớp 8: Tính
a,$\frac{3x-1}{x+5}$ + $\frac{2x+6}{x+5}$
b, 3x/x^2-4 + 2/x-2
c, x+3/x^2-6x+9 + 5/x-3
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: Tính
a,$\frac{3x-1}{x+5}$ + $\frac{2x+6}{x+5}$
b, 3x/x^2-4 + 2/x-2
c, x+3/x^2-6x+9 + 5/x-3
Comments ( 2 )
a)\dfrac{{3x – 1}}{{x + 5}} + \dfrac{{2x + 6}}{{x + 5}} = \dfrac{{3x – 1 + 2x + 6}}{{x + 5}} = \dfrac{{5x + 5}}{{x + 5}}\\
b)\dfrac{{3x}}{{{x^2} – 4}} + \dfrac{2}{{x – 2}}\\
= \dfrac{{3x}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} + \dfrac{2}{{x – 2}}\\
= \dfrac{{3x}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} + \dfrac{{2x + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{3x + 2x + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{5x + 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
c)\dfrac{{x + 3}}{{{x^2} – 6x + 9}} + \dfrac{5}{{x – 3}}\\
= \dfrac{{x + 3}}{{{{\left( {x – 3} \right)}^2}}} + \dfrac{5}{{x – 3}}\\
= \dfrac{{x + 3}}{{{{\left( {x – 3} \right)}^2}}} + \dfrac{{5x – 15}}{{{{\left( {x – 3} \right)}^2}}}\\
= \dfrac{{x + 3 + 5x – 15}}{{{{\left( {x – 3} \right)}^2}}}\\
= \dfrac{{6x – 12}}{{{{\left( {x – 3} \right)}^2}}}
\end{array}\]