\(a,x\left(x-3\right)=x^2-6\\ \Rightarrow x^2-3x-x^2=-6\\ \Rightarrow-3x=-6\\ \Rightarrow x=2\\ b,x^2-7x+12=0\\ \Rightarrow\left(x^2-3x\right)-\left(4x-12\right)=0\\ \Rightarrow x\left(x-3\right)-4\left(x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ d,x^3-25x=0\\ \Rightarrow x\left(x^2-25\right)=0\\ \Rightarrow x\left(x-5\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x0=\\x=5\\x=-5\end{matrix}\right.\)

a 2x-3=4x+6
(=) 2x-4x=6+3
(=)-2x=9
(=)x=-\(\dfrac{9}{2}\)

c: =>x(x-1+x+3)=0

=>x(2x+2)=0

=>x=-1 hoặc x=0

d: =>x(x+1)-x(x-3)=4x

=>x^2+x-x^2+3x=4x

=>4x=4x(luôn đúng)

\(a,\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow x^2=-1\left(vô.lí\right)\Rightarrow x\in\varnothing\\ c,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\\ d,\Rightarrow x^2=3\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

a) \(\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

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

b) \(x^2+1=0\)

\(\Rightarrow x^2=-1\left(VLý.do.x^2\ge0\forall x\right)\)

Vậy \(S=\varnothing\)

c) \(\Rightarrow x=\pm\sqrt{2}\)

d) \(\Rightarrow x^2=3\Rightarrow x=\pm\sqrt{3}\)

\(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Rightarrow A\)

c: Ta có: \(x^3+3x^2+3x-7=0\)

\(\Leftrightarrow x+1=2\)

hay x=1

b: Ta có: \(x\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

hay x=1/2

b: \(\Leftrightarrow\left(3x-1\right)^2=0\)

hay x=1/3

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

hay x=-4

a) ⇒ \(\left(x-\dfrac{1}{2}\right)^2\)= 0

⇒ \(x-\dfrac{1}{2}=0\)

⇒ \(x=\dfrac{1}{2}\)

b) ⇒ \(\left(3x-1\right)^2=0\)

⇒ \(3x-1=0\)

⇒ \(3x=1\)

⇒ \(x=\dfrac{1}{3}\)

c) ⇒ \(\left(x+4\right)^2=0\)

⇒ \(x+4=0\)

⇒ \(x=-4\)

quy

đồng

mà

cx

ko bt làm á

\(A=\left(\dfrac{x+8}{x\sqrt{x}+8}-\dfrac{2}{x-2\sqrt{x}+4}\right):\dfrac{1}{\sqrt{x}-1}\left(ĐKXĐ:x\ge0;x\ne1\right)\)

\(=\left[\dfrac{x+8}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}-\dfrac{2}{x-2\sqrt{x}+4}\right].\left(\sqrt{x}-1\right)\)

\(=\dfrac{x+8-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\left(\sqrt{x}-1\right)\)

\(=\dfrac{x+8-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\left(\sqrt{x}-1\right)\)

\(=\dfrac{x-2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\left(\sqrt{x}-1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

Vậy \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\) , với  \(x\ne1;x\ge0\)

\(A=\left(\dfrac{x+8}{\left(\sqrt{x}\right)^3+8}-\dfrac{2}{x-2\sqrt{x}+4}\right):\dfrac{1}{\sqrt{x}-1}\\ =\dfrac{x+8-2.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\times\dfrac{\sqrt{x}-1}{1}\\ =\dfrac{x+8-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\dfrac{\sqrt{x}-1}{1}\\ =\dfrac{x-2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\left(\sqrt{x}-1\right)\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)