Toán Lớp 8: bài 1: cộng các phân thức sau:
a,$\frac{5x+3}{x^2-3x}$+ $\frac{9-x}{9-3x}$
b, $\frac{6x^2+11x+4}{x^3-1}$+ $\frac{2x-1}{x^2+x+1}$+ $\frac{7}{1-x}$
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: bài 1: cộng các phân thức sau:
a,$\frac{5x+3}{x^2-3x}$+ $\frac{9-x}{9-3x}$
b, $\frac{6x^2+11x+4}{x^3-1}$+ $\frac{2x-1}{x^2+x+1}$+ $\frac{7}{1-x}$
Comments ( 1 )
a)\dfrac{{5x + 3}}{{{x^2} – 3x}} + \dfrac{{9 – x}}{{9 – 3x}}\\
= \dfrac{{5x + 3}}{{x\left( {x – 3} \right)}} + \dfrac{{x – 9}}{{3\left( {x – 3} \right)}}\\
= \dfrac{{3\left( {5x + 3} \right) + \left( {x – 9} \right).x}}{{3x\left( {x – 3} \right)}}\\
= \dfrac{{15x + 9 + {x^2} – 9x}}{{3x\left( {x – 3} \right)}}\\
= \dfrac{{{x^2} + 6x + 9}}{{3x\left( {x – 3} \right)}}\\
= \dfrac{{{{\left( {x + 3} \right)}^2}}}{{3x\left( {x – 3} \right)}}\\
b)\dfrac{{6{x^2} + 11x + 4}}{{{x^3} – 1}} + \dfrac{{2x – 1}}{{{x^2} + x + 1}} + \dfrac{7}{{1 – x}}\\
= \dfrac{{6{x^2} + 11x + 4 + \left( {2x – 1} \right)\left( {x – 1} \right) – 7\left( {{x^2} + x + 1} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{6{x^2} + 11x + 4 + 2{x^2} – 2x – x + 1 – 7{x^2} – 7x – 7}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{{x^2} + x – 2}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{\left( {x – 1} \right)\left( {x + 2} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{x + 2}}{{{x^2} + x + 1}}
\end{array}$