Cho `A=({x+1}/{1-x}-{1-x}/{x+1}-{4x^2}/{x^2-1}):{4x^2-4}/{x^2-2x+1}` `a).` Rút gọn `A` `b).` Tìm `x` để `A` nguyên `c).` Tìm

2 bình luận về “Cho `A=({x+1}/{1-x}-{1-x}/{x+1}-{4x^2}/{x^2-1}):{4x^2-4}/{x^2-2x+1}` `a).` Rút gọn `A` `b).` Tìm `x` để `A` nguyên `c).` Tìm”

1. $\color{Orange}{\text{~Orange~}}$

   a,

   x^2-1=-(1-x)(1+x)

   4x^2-4=4(x^2-1)=4(x-1)(x+1)

   x^2-2x+1=(x-1)^2

   A=((x+1)/(1-x)-(1-x)/(x+1)-(4x^2)/(x^2-1)):(4x^2-4)/(x^2-2x+1) (đk: x ne +-1)

   =((x+1)/(1-x)-(1-x)/(x+1)+(4x^2)/((1-x)(1+x)))*(x-1)^2/(4(x-1)(x+1))

   =((x+1)^2-(1-x)^2+4x^2)/((1-x)(1+x))*(x-1)/(4(x+1))

   =(x^2+2x+1-(1-2x+x^2)+4x^2)/((1-x)(1+x))*(x-1)/(4(x+1))

   =(x^2+2x+1-1+2x-x^2+4x^2)/((1-x)(1+x))*(x-1)/(4(x+1))

   =(4x^2+4x)/((1-x)(1+x))*(x-1)/(4(x+1))

   =(4x(x+1))/((1-x)(1+x))*(x-1)/(4(x+1))

   =(4x)/(1-x)*(x-1)/(4(x+1))

   =(-x)/(x+1)

   Vậy A=(-x)/(x+1) với x ne +-1

   b, A=(-x)/(x+1)

   =(-x-1+1)/(x+1)

   =(-(x+1)+1)/(x+1)

   =-1+1/(x+1)

   Để A nguyên thì

   1 vdots (x+1)

   ⇒(x+1) inƯ(1)+{-1;1}

   +) x+1=1⇒x=0 (tmđk)

   +) x+1=-1⇒x=-2 (tmđk)

   Vậy x in {-2;0} thì A nguyên

   A>1

   ⇒A-1>0

   ⇒(-x)/(x+1)-1>0

   ⇒(-x-(x+1))/(x+1)>0

   ⇒(-x-x-1)/(x+1)>0

   ⇒(-2x-1)/(x+1)>0

   ⇔$\left[\begin{matrix} \begin{cases} -2x-1>0\\x+1>0 \end{cases}\\ \begin{cases} -2x-1<0\\x+1<0\end{cases}\end{matrix}\right.$

   ⇔$\left[\begin{matrix} \begin{cases} -2x>1\\x>-1 \end{cases}\\ \begin{cases} -2x<1\\x<-1\end{cases}\end{matrix}\right.$

   ⇔$\left[\begin{matrix} \begin{cases} x<-\dfrac{1}{2}\\x>-1\end{cases}\\ \begin{cases} x>-\dfrac{1}{2}\\x<-1\end{cases}\end{matrix}\right.$

   ⇒-1<x<-1/2

   Vậy A>1 khi -1<x<-1/2

   Trả lời
2. a, ĐKXĐ: x \ne +-1

   Với x \ne +-1 thì

   Ta có: A = ((x + 1)/(1 – x) – (1 – x)/(x + 1) – (4x^2)/(x^2 – 1)) : (4x^2 – 4)/(x^2 – 2x + 1)

   => A = ((-x – 1)/(x – 1) + (x – 1)/(x + 1) – (4x^2)/((x – 1)(x + 1))): (4(x^2 – 1))/((x – 1)^2)

   => A = (((-x – 1)(x + 1))/((x – 1)(x + 1)) + ((x – 1)^2)/((x – 1)(x + 1)) – (4x^2)/((x – 1)(x + 1))): (4(x – 1)(x + 1))/((x – 1)^2)

   => A = (-x^2 – x -x-1 + x^2 – 2x + 1 – 4x^2)/((x – 1)(x + 1)) : (4(x + 1))/(x – 1))

   => A = (-4x^2 – 4x)/((x – 1)(x + 1)) . (x – 1)/(4(x + 1))

   => A = (-4x(x + 1))/((x – 1)(x + 1)) . (x – 1)/(4(x + 1))

   => A = (-x)/(x + 1)

   Vậy A = (-x)/(x + 1) với x \ne +-1

   b,

   Để A nguyên thì:

   -x \vdots x + 1

   => -x -1 + 1 \vdots x + 1

   => -(x + 1) + 1 \vdots x + 1

   Vì -(x + 1) \vdots x + 1

   => 1 \vdots x + 1

   => x + 1 \in Ư(1)

   => x + 1 \in {1,-1}

   => x \in {0,-2} (thỏa mãn)

   Vậy x \in {0,-2} thì A nguyên

   c,

   Để A > 1 thì

   (-x)/(x + 1) > 1

   <=> (-x)/(x + 1) – 1 > 0

   <=> (-x )/(x + 1) – (x + 1)/(x + 1) > 0

   <=> (-x – x – 1)/(x + 1) > 0

   <=> (-2x – 1)/(x + 1) > 0

   <=> $\left[\begin{matrix} \begin{cases} -2x – 1>0\\x + 1 > 0 \end{cases}\\ \begin{cases} -2x – 1 < 0\\x + 1 < 0 \end{cases}\end{matrix}\right.$ 

   <=> $\left[\begin{matrix} \begin{cases} x< \dfrac{-1}{2}\\x > -1 \end{cases}\\ \begin{cases} x > \dfrac{-1}{2}\\x < -1\end{cases}\end{matrix}\right.$ 

   <=>$\left[\begin{matrix} -1 < x < \dfrac{-1}{2}\\ \dfrac{-1}{2} ​

   <=> -1 <x < (-1)/2

   Vậy -1 < x < (-1)/2 thì A > 1

   $#duong612009$

   Trả lời