1:
a: \(A=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}=\dfrac{-4}{x+2}\)
b: 2x^2+x=0
=>x=0(loại) hoặc x=-1/2(nhận)
Khi x=-1/2 thì \(A=\dfrac{-4}{-\dfrac{1}{2}+2}=-4:\dfrac{3}{2}=\dfrac{-8}{3}\)
c: A=1/2
=>x+2=-8
=>x=-10
d: A nguyên dương
=>x+2<0 và \(x+2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{-3;-4;-6\right\}\)
Câu 1:
a. \(A=\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\right).\left(\dfrac{2}{x}-1\right)\left(đkxđ:x\ne0;x\ne2\right)\)
\(=\left(\dfrac{x+2}{x^2-4}+\dfrac{2x}{x^2-4}-\dfrac{x-2}{x^2-4}\right).\left(\dfrac{2}{x}-1\right)\)
\(=\dfrac{x+2+2x-x+2}{x^2-4}.\left(\dfrac{2}{x}-1\right)\)
\(=\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}.\dfrac{2-x}{x}\)
\(=\dfrac{2}{x-2}.\dfrac{x-2}{-x}\)
\(=-\dfrac{2}{x}\)
b.
\(2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)
Thay \(x=-\dfrac{1}{2}\) vào A, ta được:
\(A=-\dfrac{2}{-\dfrac{1}{2}}=4\)
c.
\(A=\dfrac{1}{2}\)
\(\Leftrightarrow-\dfrac{2}{x}=\dfrac{1}{2}\)
\(\Leftrightarrow x=-4\)
d.
\(A\in N\left(A\ne0,vì.tử.khác.0\right)\)
\(\Rightarrow x< 0;x\inƯ_{\left(2\right)}\)
\(\Rightarrow x\in\left\{-1;-2\right\}\)
2
b. Ta có: a < b
-3a > -3b (nhân cả 2 vế cho -3)
-3a + 1 > -3b + 1 (cộng cả 2 vế cho 1)