Question

Cho biểu thức N = \(\left(\dfrac{1}{y-1}-\dfrac{y}{1-y^3}.\dfrac{y^2+y+1}{y+1}\right):\dfrac{1}{y^2-1}\)

a) Rút gọn N

b) Tính giá trị của N khi y = \(\dfrac{1}{2}\)

c) Tìm giá trị của y để N luôn có giá trị dương

Answer 1

a)\(N=\left(\dfrac{1}{y-1}-\dfrac{y}{1-y^3}.\dfrac{y^2+y+1}{y+1}\right):\dfrac{1}{y^2-1}\)

\(N=\left(\dfrac{1}{y-1}+\dfrac{y}{y^3-1}.\dfrac{y^2+y+1}{y+1}\right).\left(y^2-1\right)\)

\(N=\left(\dfrac{1}{y-1}+\dfrac{y}{\left(y-1\right)\left(y^2+y+1\right)}.\dfrac{y^2+y+1}{y+1}\right).\left(y^2-1\right)\)

\(N=\left(\dfrac{1}{y-1}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}\right).\left(y^2-1\right)\)

\(N=\left(\dfrac{1}{y-1}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}\right).\left(y-1\right).\left(y+1\right)\)

\(N=y+1+y\)

N=2y+1

b) Khi y=1/2 thì N=\(2.\dfrac{1}{2}+1=1+1=2\)

c) N dương <=> N=2y+1>0 <=> 2y>-1 <=> y>-1/2