Rút gọn bt dùm mk vs ạ!
a) X²+2/x+3 – 5/x²+x-6 + 1/2-x
b) (a+3)²/2a²+6a.(1-6a-18/a²-9)
c) x/2x-2+x²+1/2-2x²
d) 2x²-4x+8/x³+8
e) x²-4x+4/x²-4
Rút gọn bt dùm mk vs ạ!
a) X²+2/x+3 – 5/x²+x-6 + 1/2-x
b) (a+3)²/2a²+6a.(1-6a-18/a²-9)
c) x/2x-2+x²+1/2-2x²
d) 2x²-4x+8/x³+8
e) x²-4x+4/x²-4
Đáp án:
\(e)\dfrac{{x – 2}}{{x + 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{{x^2} + 2}}{{x + 3}} – \dfrac{5}{{\left( {x + 3} \right)\left( {x – 2} \right)}} + \dfrac{1}{{2 – x}}\\
= \dfrac{{\left( {{x^2} + 2} \right)\left( {x – 2} \right) – 5 – x – 3}}{{\left( {x + 3} \right)\left( {x – 2} \right)}}\\
= \dfrac{{{x^3} – 2{x^2} + 2x – 4 – x – 8}}{{\left( {x + 3} \right)\left( {x – 2} \right)}}\\
= \dfrac{{{x^3} – 2{x^2} + x – 12}}{{\left( {x + 3} \right)\left( {x – 2} \right)}}\\
b)\dfrac{{{{\left( {a + 3} \right)}^2}}}{{2{a^2} + 6a}}.\left( {1 – \dfrac{{6a – 18}}{{{a^2} – 9}}} \right)\\
= \dfrac{{a + 3}}{{2a}}.\dfrac{{{a^2} – 9 – 6a + 18}}{{\left( {a + 3} \right)\left( {a – 3} \right)}}\\
= \dfrac{{a + 3}}{{2a}}.\dfrac{{{a^2} – 6a + 9}}{{\left( {a + 3} \right)\left( {a – 3} \right)}}\\
= \dfrac{{{{\left( {a – 3} \right)}^2}}}{{2a\left( {a – 3} \right)}} = \dfrac{{a – 3}}{{2a}}\\
c)\dfrac{x}{{2x – 2}} + \dfrac{{{x^2} + 1}}{{2 – 2{x^2}}}\\
= \dfrac{{x\left( {x + 1} \right) – {x^2} – 1}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x – 1}}{{2\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{1}{{2\left( {x + 1} \right)}}\\
d)\dfrac{{2{x^2} – 4x + 8}}{{\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right)}}\\
e)\dfrac{{{{\left( {x – 2} \right)}^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{{x – 2}}{{x + 2}}
\end{array}\)