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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 \(x\in\varnothing\)
\(a,ĐK:x\ne1;x\ne-1\\ b,C=\dfrac{x^2+x+x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+2x+1}{2x^2-2}\\ c,C=-\dfrac{1}{2}\Leftrightarrow2-2x^2=2x^2+2x+1\\ \Leftrightarrow4x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-1}{4}\\x=\dfrac{-\sqrt{5}-1}{4}\end{matrix}\right.\\ d,C>0\Leftrightarrow2x^2-2>0\left(2x^2+2x+1>0\right)\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
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: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
