\(\left(-4x^2zy^4\right)\left(-\dfrac{1}{2}zx^2\right)^2\\ =-4x^2zy^4.\dfrac{1}{4}z^2x^4\\ =\left(-4.\dfrac{1}{4}\right)\left(x^2.x^4\right).y^4\left(z.z^2\right)\\ =-x^6y^4z^3\)

\(Th1:a\le\dfrac{1}{2}\) nhé

`sqrt{1-4a+4a^2}-2a`

`=sqrt{(1-2a)^2}-2a`

`=|1-2a|-2a`

`TH1: a <= 1/2`

`=>1-2a >= 0`

`=> |1-2a|=1-2a`

Ta có:

`|1-2a|-2a=1-2a-2a=1-4a`

`TH2:a > 1/2`

`=> 1-2a<0`

`=>|1-2a|=-(1-2a)=2a-1`

Ta có:

`|1-2a|-2a=2a-1-2a=-1`

\(K=2\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a^2-a}\\ =2\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\dfrac{a\left(a-1\right)}{\sqrt{a}+1}\\ =2.\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\sqrt{a}.\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\\ =2\sqrt{a}\)

`2, K= sqrt2012`

`<=>2sqrt{a}=sqrt2012`

`<=>sqrt4 . sqrt{a}=sqrt2012`

`<=>sqrt{a}=503`

`<=>a=503(tm)`

xem lại đi .-.

\(ĐKXĐ:x\ne1\\ \dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\\ \Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2-2x^2+2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\\ \Leftrightarrow\dfrac{-4x^2+3x+1}{\left(x-1\right)\left(x^2+x+1\right)}=0\\ \Rightarrow4x^2-3x-1=0\)

`<=>4x^2 -4x+x-1=0`

`<=>4x(x-1)+(x-1)=0`

`<=>(4x+1)(x-1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy `S={-1/4}`

`x^2 -5x+6=0`

`<=>x^2 -2x-3x+6=0`

`<=>x(x-2)-3(x-2)=0`

`<=>(x-2)(x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)