bài 1 :thực hiện các phép tính sau
a,11x-4/x-1 + 10x+4/2-2x
b,-3x mũ 2/x mũ 3+1 + 1/x mũ 2-x+1+1/x+1
c,1/x+2 + 5/2x mũ 2+3x-2
d,1/xy-y mũ 2 – 1/x mũ 2-xy
bài 1 :thực hiện các phép tính sau
a,11x-4/x-1 + 10x+4/2-2x
b,-3x mũ 2/x mũ 3+1 + 1/x mũ 2-x+1+1/x+1
c,1/x+2 + 5/2x mũ 2+3x-2
d,1/xy-y mũ 2 – 1/x mũ 2-xy
Đáp án:
a) 6
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{11x – 4}}{{x – 1}} + \dfrac{{10x + 4}}{{2\left( {1 – x} \right)}}\\
= \dfrac{{22x – 8 – 10x – 4}}{{2\left( {x – 1} \right)}}\\
= \dfrac{{12x – 12}}{{2\left( {x – 1} \right)}} = \dfrac{{12}}{2} = 6\\
b)\dfrac{{ – 3{x^2}}}{{{x^3} + 1}} + \dfrac{1}{{{x^2} – x + 1}} + \dfrac{1}{{x + 1}}\\
= \dfrac{{ – 3{x^2} + x + 1 + {x^2} – x + 1}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
= \dfrac{{ – 2{x^2} + 2}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = \dfrac{{ – 2\left( {x + 1} \right)\left( {x – 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}}\\
= \dfrac{{ – 2\left( {x – 1} \right)}}{{{x^2} – x + 1}}\\
c)\dfrac{1}{{x + 2}} + \dfrac{5}{{2{x^2} + 3x – 2}}\\
= \dfrac{{2x – 1 + 5}}{{\left( {x + 2} \right)\left( {2x – 1} \right)}}\\
= \dfrac{{2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {2x – 1} \right)}} = \dfrac{2}{{2x – 1}}\\
d)\dfrac{1}{{xy – {y^2}}} – \dfrac{1}{{{x^2} – xy}}\\
= \dfrac{1}{{y\left( {x – y} \right)}} – \dfrac{1}{{x\left( {x – y} \right)}}\\
= \dfrac{{x – y}}{{xy\left( {x – y} \right)}} = \dfrac{1}{{xy}}
\end{array}\)