b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

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 \(x\in\varnothing\)

\(a,ĐK:x\ne\pm1\\ b,B=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\\ c,B=-\dfrac{1}{2}\Leftrightarrow2\left(x+1\right)=-2\Leftrightarrow x+1=-1\Leftrightarrow x=-2\left(tm\right)\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a, \(A=x^2\left(2x-1\right)+x\left(x+8\right)=2x^3-x^2+x^2+8x=2x^3+8x\)

Thay x = -2, ta có:

\(2\cdot\left(-2\right)^3+8\cdot\left(-2\right)=-32\)

b, \(A=2x^3+8x=0\\ \Leftrightarrow2x\left(x^2+4\right)=0\\ \Leftrightarrow x=0\)

Vậy A=0 khi x=0

a,A = \(x^2\).( 2\(x\) - 1) + \(x\)(\(x+8\))

A = 2\(x^3\) - \(x^2\) + \(x^2\) + 8\(x\)

A = 2\(x^3\) + 8\(x\)

b, \(x=-2\) ⇒ A = 2.(-2)³ + 8.(-2) = - 32 

A = 0 ⇔ 2\(x^3\) + 8\(x\) = 0

             2\(x\left(x^2+4\right)\) = 0 

              vì \(x^2\) + 4 > 0 ∀ \(x\) ⇒ \(x\)  =0 

\(a,A=x^2\left(2x-1\right)+x\left(x+8\right)\\ =2x^3-x^2+x^2+8x=2x^3+8x\\ =2x.\left(x^2+8\right)\\ Thế.x=-2.vào.A:A=2x.\left(x^2+8\right)=2.\left(-2\right).\left[\left(-2\right)^2+8\right]=-4.12=-48\\ b,A=0\\ \Leftrightarrow2x\left(x^2+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+8=0\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=0\\ Vậy:A=0\Leftrightarrow x=0\)