Toán Lớp 8: bài 2: cộng các phân thức đại số
a,$\frac{6x}{x^3+1}$+ $\frac{1}{x^2-x+1}$ +$\frac{2}{x+1}$
b,$\frac{2}{(x-y)(y-z)}$+ $\frac{2}{(z-x)(x-y)}$+ $\frac{2}{(y-z)(z-x)}$
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 2: cộng các phân thức đại số
a,$\frac{6x}{x^3+1}$+ $\frac{1}{x^2-x+1}$ +$\frac{2}{x+1}$
b,$\frac{2}{(x-y)(y-z)}$+ $\frac{2}{(z-x)(x-y)}$+ $\frac{2}{(y-z)(z-x)}$
Comments ( 1 )
a)\dfrac{{6x}}{{{x^3} + 1}} + \dfrac{1}{{{x^2} – x + 1}} + \dfrac{2}{{x + 1}}\\
= \dfrac{{6x + x + 1 + 2\left( {{x^2} – x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
= \dfrac{{7x + 1 + 2{x^2} – 2x + 2}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
= \dfrac{{2{x^2} + 5x + 3}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
= \dfrac{{\left( {x + 1} \right)\left( {2x + 3} \right)}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
= \dfrac{{2x + 3}}{{{x^2} – x + 1}}\\
b)\dfrac{2}{{\left( {x – y} \right)\left( {y – z} \right)}} + \dfrac{2}{{\left( {z – x} \right)\left( {x – y} \right)}} + \dfrac{2}{{\left( {y – z} \right)\left( {z – x} \right)}}\\
= \dfrac{{2\left( {z – x} \right) + 2\left( {y – z} \right) + 2\left( {x – y} \right)}}{{\left( {x – y} \right)\left( {y – z} \right)\left( {z – x} \right)}}\\
= \dfrac{{2z – 2x + 2y – 2z + 2x – 2y}}{{\left( {x – y} \right)\left( {y – z} \right)\left( {z – x} \right)}}\\
= 0
\end{array}$