`a, Q?`
b,`
\(B=\dfrac{x^2+4}{x^2-4}-\dfrac{2}{x-2}\\ =\dfrac{x^2+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}\\ =\dfrac{x^2+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+4-2x-4}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x}{x+2}\left(dpcm\right)\)
`c,` Ta có `A-B<3/2`
\(\Rightarrow\dfrac{3x}{x+2}-\dfrac{x}{x+2}< \dfrac{3}{2}\\ \Leftrightarrow\dfrac{3x-x}{x+2}< \dfrac{3}{2}\\ \Leftrightarrow\dfrac{2\left(3x-x\right)}{2\left(x+2\right)}< \dfrac{3\left(x+2\right)}{2\left(x+2\right)}\\ \Leftrightarrow6x-2x< 3x+6\\\Leftrightarrow6x-2x-3x< 6\\ \Leftrightarrow x< 6 \)
Mà `x >= -2`
`->`Số dương lớn nhất thoả mãn là `5`
bổ sung câu hỏi
c) Với x > -2, tìm số nguyên dương x lớn nhất thỏa mãn A - B < \(\dfrac{3}{2}\)
a: Khi x=9 thì \(A=\dfrac{3\cdot9}{9+2}=\dfrac{27}{11}\)
b: \(B=\dfrac{x^2+4-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
c: A-B<3/2
=>\(\dfrac{2x}{x+2}-\dfrac{3}{2}< 0\)
=>\(\dfrac{4x-3x-6}{2\left(x+2\right)}< 0\)
=>\(\dfrac{x-6}{x+2}< 0\)
=>-2<x<6
mà x>-2
nên số nguyên dương x lớn nhất thỏa mãn là x=5
