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\(a,ĐK:x\ne1;x\ne-1\\ b,C=\dfrac{x^2+x+x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+2x+1}{2x^2-2}\\ c,C=-\dfrac{1}{2}\Leftrightarrow2-2x^2=2x^2+2x+1\\ \Leftrightarrow4x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-1}{4}\\x=\dfrac{-\sqrt{5}-1}{4}\end{matrix}\right.\\ d,C>0\Leftrightarrow2x^2-2>0\left(2x^2+2x+1>0\right)\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Câu b rút gọn C sai rồi, phải là \(\dfrac{1}{2\left(x+1\right)}\) chứ.
\(a,ĐK:x\ne\pm2\\ b,A=\dfrac{5x+10+14x-28-20}{2\left(x-2\right)\left(x+2\right)}=\dfrac{19\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}=\dfrac{19}{2\left(x+2\right)}\\ c,x=-\dfrac{1}{2}\Leftrightarrow A=\dfrac{19}{2\left(2-\dfrac{1}{2}\right)}=\dfrac{19}{2\cdot\dfrac{3}{2}}=\dfrac{19}{3}\)
a: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{1}{2};-2\right\}\)
b: \(B=\dfrac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x+2}\)
\(=\dfrac{8x-4}{2x-1}\cdot\dfrac{1}{x+2}=\dfrac{4}{x+2}\)
a, Để C có nghĩa thì \(\hept{\begin{cases}2x-2\ne0\\2-2x\ne0\end{cases}\Rightarrow}x\ne1\)
b, Với x khác 1 thì
\(C=\frac{x}{2x-2}+\frac{x^2+1}{2-2x}=\frac{-x}{2-2x}+\frac{x^2+1}{2-2x}=\frac{x^2-x+1}{2-2x}\)
c, \(C=-0,5\Rightarrow\frac{x^2-x+1}{2-2x}=\frac{-1}{2}\)
\(\Rightarrow2\left(x^2-x+1\right)=\left(2-2x\right).\left(-1\right)\)
\(\Rightarrow2x^2-2x+2=-2+2x\)
\(\Rightarrow2x^2-2x+2+2-2x=0\)
\(\Rightarrow2x^2-4x+4=0\Rightarrow2\left(x^2-2x+2\right)=0\)
\(x^2-2x+2=\left(x-1\right)^2+1>0\forall x\)
Do đó: \(2\left(x^2-2x+2\right)>0\forall x\)
Vậy \(x\in\varnothing\)
a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)
b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)
\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)
c: Thay x=2017 vào C, ta được:
\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)
