Cho A=($\frac{3+x}{3-x}$ – $\frac{3-x}{x+3}$ +$\frac{4x^2}{x^2 -9}$).(2+x . $\frac{x^2 + x+1}{8-x}$)
Rút gọn A
Giúp mk với cần gấp
Cho A=($\frac{3+x}{3-x}$ – $\frac{3-x}{x+3}$ +$\frac{4x^2}{x^2 -9}$).(2+x . $\frac{x^2 + x+1}{8-x}$) Rút gọn A Giúp mk với cần gấp
By Peyton
Đáp án:
\({\dfrac{{\left( { – 2{x^2} + 12x} \right)\left( {{x^3} + 3{x^2} + 3x + 2} \right)}}{{\left( {9 – {x^2}} \right)\left( {8 – x} \right)}}}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{DK:x \ne \pm 3;x \ne 8}\\
{A = \left( {\dfrac{{3 + x}}{{3 – x}} – \dfrac{{3 – x}}{{x + 3}} + \dfrac{{4{x^2}}}{{{x^2} – 9}}} \right).\left( {\left( {2 + x} \right).\dfrac{{{x^2} + x + 1}}{{8 – x}}} \right)}\\
{ = \left[ {\dfrac{{{x^2} + 6x + 9 – 9 + 6x – {x^2} – 4{x^2}}}{{\left( {3 – x} \right)\left( {x + 3} \right)}}} \right].\left[ {\dfrac{{2{x^2} + 2x + 2 + {x^3} + {x^2} + x}}{{8 – x}}} \right]}\\
{ = \dfrac{{ – 2{x^2} + 12x}}{{\left( {3 – x} \right)\left( {x + 3} \right)}}.\dfrac{{{x^3} + 3{x^2} + 3x + 2}}{{8 – x}}}\\
{ = \dfrac{{\left( { – 2{x^2} + 12x} \right)\left( {{x^3} + 3{x^2} + 3x + 2} \right)}}{{\left( {9 – {x^2}} \right)\left( {8 – x} \right)}}}
\end{array}\)
\(\begin{array}{l}
b.2x = 4\\
\to x = 2\\
\to A = \dfrac{{\left( { – 2.4 + 12.2} \right)\left( {8 + 3.4 + 3.2 + 2} \right)}}{{\left( {9 – 4} \right)\left( {8 – 2} \right)}}\\
= \dfrac{{16.28}}{{5.6}} = \dfrac{{224}}{{15}}
\end{array}\)