a: \(Q=1+\dfrac{1}{x+2}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{x+2}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{1}{x+2}:\dfrac{4x+8-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{1}{x+2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x+10}\)
\(=1+\dfrac{x-2}{2x+10}\)
\(=\dfrac{2x+10+x-2}{2x+10}=\dfrac{3x+8}{2x+10}\)
b: Để Q=1 thì 3x+8=2x+10
hay x=2(loại)
Để Q=0 thì 3x+8=0
hay x=-8/3(nhận)
a) \(Q=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right).\)
\(\left(đkxđ:x\ne-2;x\ne2;x\ne-3;x\ne0\right).\)
\(Q=1+\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right).\)
\(Q=1+\dfrac{1}{x+2}:\left(\dfrac{2}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right).\)
\(Q=1+\dfrac{1}{x+2}:\dfrac{2\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}.\)
\(Q=1+\dfrac{1}{x+2}.\dfrac{\left(x-2\right)\left(x+2\right)}{2x+4-x-x+2}.\)
\(Q=1+\dfrac{x-2}{6}=\dfrac{6+x-2}{6}=\dfrac{x+4}{6}.\)
b) \(Q=1.\Rightarrow\dfrac{x+4}{6}=1.\Leftrightarrow x+4=6.\Leftrightarrow x=2\left(koTM\right).\)
\(Q=0.\Rightarrow\dfrac{x+4}{6}=0.\Leftrightarrow x+4=0.\Leftrightarrow x=-4\left(TM\right).\)
c) \(\left|x-3\right|=1.\Leftrightarrow\left[{}\begin{matrix}x-3=1.\\x-3=-1.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(TM\right).\\x=2\left(koTM\right).\end{matrix}\right.\)
Thay \(x=4\) vào biểu thức \(Q\): \(\dfrac{4+6}{6}=\dfrac{10}{6}=\dfrac{5}{3}.\)
