`1)` Biểu thức xác định `<=>x+1 \ne 0<=>x \ne -1`
`[x^2+2x+1]/[x+1]=[(x+1)^2]/[x+1]=x+1`
`2)` Bth xác định `<=>x(x-3) \ne 0<=>{(x \ne 0),(x \ne 3):}`
`[x^2-6x+9]/[x(x-3)]=[(x-3)^]/[x(x-3)]=[x-3]/x`
`3)` Bth xác định `<=>2x(x+2) \ne 0<=>{(x \ne 0),(x \ne -2):}`
`[x^2-4]/[2x(x+2)]=[(x-2)(x+2)]/[2x(x+2)]=[x-2]/[2x]`
`4)` Bth xác định `<=>5x^2-10x \ne 0<=>5x(x-2) \ne 0<=>{(x \ne 0),(x \ne 2):}`
`[x^2-2x]/[5x^2-10x]=[x(x-2)]/[5x(x-2)]=1/5`
1)
\(ĐKXĐ:x\ne-1\)
\(\dfrac{x^2+2x+1}{x+1}\\ =\dfrac{\left(x+1\right)^2}{x+1}\\ =x+1\)
2)
ĐKXĐ x khác 0 và x khác 3
\(\dfrac{x^2-6x+9}{x\left(x-3\right)}\\ =\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}\\ =\dfrac{x-3}{x}\)
3)
ĐKXĐ: x khác 0 và x khác -2
\(\dfrac{x^2-4}{2x\left(x+2\right)}\\ =\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)}\\ =\dfrac{x-2}{2x}\)
4)
DKXĐ: x khác 0 và x khác 2
\(\dfrac{x^2-2x}{5x^2-10x}\\ =\dfrac{x\left(x-2\right)}{5x\left(x-2\right)}\\ =\dfrac{1}{5}\)
đk `x≠-1`
`(x^2+2x+1)/(x+1)`
`=((x+1)^2)/(x+1)`
`=x+1`
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đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)
`(x^2-6x+9)/(x(x-3))`
`=((x-3)^2)/(x(x-3))`
`=(x-3)/x`
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đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-2\end{matrix}\right.\)
`(x^2-4)/(2x(x+2))`
`=((x-2)(x+2))/(2x(x+2))`
`=(x-2)/(2x)`
--------
đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
`(x^2-2x)/(5x^2-10x)`
`=(x(x-2))/(5x(x-2))`
`=x/(5x)`
\(1,ĐKXĐ:x\ne-1\)
\(\dfrac{x^2+2x+1}{x+1}=\dfrac{\left(x+1\right)^2}{x+1}=x+1\)
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\(2,ĐKXĐ:x\ne0;3\)
\(\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
_________________________________
\(3,ĐKXĐ:x\ne0;-2\)
\(\dfrac{x^2-4}{2x\left(x+2\right)}=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)}=\dfrac{x-2}{2x}\)
_________________________________
\(4,ĐKXĐ:x\ne0;2\)
\(\dfrac{x^2-2x}{5x^2-10x}=\dfrac{x\left(x-2\right)}{5x\left(x-2\right)}=\dfrac{x}{5x}=\dfrac{1}{5}\)
