Toán Lớp 8: thực hiện phép tính: (x+2/x+1 – 2x/x-1)*3x+3/x+4x^2+x+7/x^2-x
em cần gấp ạ
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: thực hiện phép tính: (x+2/x+1 – 2x/x-1)*3x+3/x+4x^2+x+7/x^2-x
em cần gấp ạ
Comments ( 1 )
DK:x \ne \left\{ { – 1;0;1} \right\}\\
\left( {\dfrac{{x + 2}}{{x + 1}} – \dfrac{{2x}}{{x – 1}}} \right).\dfrac{{3x + 3}}{x} + \dfrac{{4{x^2} + x + 7}}{{{x^2} – x}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x – 1} \right) – 2x\left( {x + 1} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{3\left( {x + 1} \right)}}{x} + \dfrac{{4{x^2} + x + 7}}{{x\left( {x – 1} \right)}}\\
= \dfrac{{{x^2} + x – 2 – 2{x^2} – 2x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{3\left( {x + 1} \right)}}{x} + \dfrac{{4{x^2} + x + 7}}{{x\left( {x – 1} \right)}}\\
= \dfrac{{ – {x^2} – x – 2}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{3\left( {x + 1} \right)}}{x} + \dfrac{{4{x^2} + x + 7}}{{x\left( {x – 1} \right)}}\\
= \dfrac{{3\left( { – {x^2} – x – 2} \right)}}{{x\left( {x – 1} \right)}} + \dfrac{{4{x^2} + x + 7}}{{x\left( {x – 1} \right)}}\\
= \dfrac{{ – 3{x^2} – 3x – 6}}{{x\left( {x – 1} \right)}} + \dfrac{{4{x^2} + x + 7}}{{x\left( {x – 1} \right)}}\\
= \dfrac{{ – 3{x^2} – 3x – 6 + 4{x^2} + x + 7}}{{x\left( {x – 1} \right)}}\\
= \dfrac{{{x^2} – 2x + 1}}{{x\left( {x – 1} \right)}}\\
= \dfrac{{{{\left( {x – 1} \right)}^2}}}{{x\left( {x – 1} \right)}} = \dfrac{{x – 1}}{x}
\end{array}\)