Question

cho \(M=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

a) Tìm ĐKXĐ và rút gọn M

b) Tìm x thuộc Z để M đạt GTLN

Answer 1

\(M=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(M=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)

\(M=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{6}{x+2}\right)\)

a) dkxd : x khac {0;1;-2)

\(M=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right).\left(\dfrac{x+2}{6}\right)\)

\(M=\left(\dfrac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-1\right)\left(x+2\right)}\right).\left(\dfrac{x+2}{6}\right)=\dfrac{-6}{6\left(x-2\right)}=\dfrac{1}{2-x}\)

b)

GTLN M =1 khi x =1