a: Khi x=-2 thì \(B=\dfrac{4+7}{-2}=\dfrac{-11}{2}\)
b: \(A=\dfrac{x^2-3x+\left(2x-1\right)\left(x+3\right)-2x^2+x+3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-x^2-2x+3+2x^2+6x-x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x}{x-3}\)
c: 1/A+B
=(x-3)/x+(x^2+7)/x=(x^2+x+4)/x
(x^2+x+4)/x<=5
=>(x^2+x+4)/x-5<=0
=>(x^2+x+4-5x)/x<=0
=>(x-2)^2/x<=0
mà x>0
nên x-2=0
=>x=2
Đúng 0
Bình luận (0)
