Toán Lớp 8: Tìm giá trị củaE=1+((2x^3+x^2-x)/(x^3-1)-(2x-1)/(x-1))*(x^2-x)/(2x-1)
Biết x^2+x-6=0
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Toán Lớp 8: Tìm giá trị củaE=1+((2x^3+x^2-x)/(x^3-1)-(2x-1)/(x-1))*(x^2-x)/(2x-1)
Biết x^2+x-6=0
Comments ( 2 )
x^2+x-6=0
<=>x^2-2x+3x-6=0
<=>(x^2-2x)+(3x-6)=0
<=>x.(x-2)+3.(x-2)=0
<=>(x+3)(x-2)=0
<=>[(x+3=0),(x-2=0):}
<=>[(x=-3),(x=2):}
E=1+((2x^3+x^2-x)/(x^3-1)-(2x-1)/(x-1)).(x^2-x)/(2x-1)( x\ne1;1/2 )
E=1+((x.(2x^2+x-1))/((x-1)(x^2+x+1))-(2x-1)/(x-1)) . (x.(x-1))/(2x-1)
E=1+((x.(2x^2+2x-x-1))/((x-1)(x^2+x+1))-(2x-1)/(x-1)).(x.(x-1))/(2x-1)
E=1+((x.[2x.(x+1)-(x+1)])/((x-1)(x^2+x+1))-(2x-1)/(x-1)).(x.(x-1))/(2x-1)
E=1+((x.(2x-1).(x+1))/((x-1)(x^2+x+1))-((2x-1).(x^2+x+1))/((x-1)(x^2+x+1))).(x.(x-1))/(2x-1)
E=1+(x.(2x-1).(x+1)-(2x-1).(x^2+x+1))/((x-1)(x^2+x+1)) . (x.(x-1))/(2x-1)
E=1+((2x-1).[x.(x+1)-x^2-x-1])/((x-1)(x^2+x+1)).(x.(x-1))/(2x-1)
E=1+((2x-1).(x^2+x-x^2-x-1))/((x-1)(x^2+x+1)) . (x.(x-1))/(2x-1)
E=1+((2x-1).(-1))/((x-1).(x^2+x+1)).(x.(x-1))/(2x-1)
E=1+(-(2x-1).x.(x-1))/((x-1)(x^2+x+1)(2x-1))
E=1+(-x)/(x^2+x+1)
E=(x^2+x+1)/(x^2+x+1)+(-x)/(x^2+x+1)
E=(x^2+1)/(x^2+x+1)
Với x=2
E=(2^2+1)/(2^2+2+1)
E=(4+1)/(4+2+1)
E=5/7
Vậy E=5/7 tại x=2
Với x=-3
E=((-3)^2+1)/((-3)^2-3+1)
E=(9+1)/(9-3+1)
E=10/7
Vậy E=10/7 tại x=-3
{E = \dfrac{{10}}{7}}\\
{E = \dfrac{5}{7}}
\end{array}} \right.\)
{a)DK:x \ne \left\{ {\dfrac{1}{2};1} \right\}}\\
{E = 1 + \left( {\dfrac{{2{x^3} + {x^2} – x}}{{{x^3} – 1}} – \dfrac{{2x – 1}}{{x – 1}}} \right).\dfrac{{{x^2} – x}}{{2x – 1}}}\\
{ = 1 + \dfrac{{2{x^3} + {x^2} – x – \left( {2x – 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{x\left( {x – 1} \right)}}{{2x – 1}}}\\
{ = 1 + \dfrac{{2{x^3} + {x^2} – x – 2{x^3} – 2{x^2} – 2x + {x^2} + x + 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{x\left( {x – 1} \right)}}{{2x – 1}}}\\
{ = 1 + \dfrac{{ – 2x + 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{x\left( {x – 1} \right)}}{{2x – 1}}}\\
{ = 1 – \dfrac{{2x – 1}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{x\left( {x – 1} \right)}}{{2x – 1}}}\\
{ = 1 – \dfrac{x}{{{x^2} + x + 1}}}\\
{ = \dfrac{{{x^2} + x + 1 – x}}{{{x^2} + x + 1}}}\\
{ = \dfrac{{{\rm{\;}}{x^2} + 1}}{{{x^2} + x + 1}}}\\
{Do:{x^2} + x – 6 = 0}\\
{ \to \left( {x + 3} \right)\left( {x – 2} \right) = 0}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = {\rm{ \;}} – 3}\\
{x = 2}
\end{array}} \right.}\\
{Thay:\left[ {\begin{array}{*{20}{l}}
{x = {\rm{ \;}} – 3}\\
{x = 2}
\end{array}} \right.}
\end{array}\)
{E = \dfrac{{{{\left( { – 3} \right)}^2} + 1}}{{{{\left( { – 3} \right)}^2} – 3 + 1}} = \dfrac{{10}}{7}}\\
{E = \dfrac{{{2^2} + 1}}{{{2^2} + 2 + 1}} = \dfrac{5}{7}}
\end{array}} \right.\)