Đáp án:
`a)` ` \sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}=2\sqrt{5}`
`b)` `\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=2\sqrt{2}`
`c)` $\dfrac{\sqrt{x^2+6x+9}}{x+3}=\begin{cases}1\ nếu\ x> -3\\-1\ nếu\ x< -3\end{cases}$
`d)` `{x-8\sqrt{x}-9}/{x-1}={\sqrt{x}-9}/{\sqrt{x}-1}` với `x\ge 0; x\ne 1`
Giải thích các bước giải:
`a)` `\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}`
`=\sqrt{5-2.\sqrt{5}.1+1^2}+\sqrt{5+2.\sqrt{5}.1+1^2}`
`= \sqrt{(\sqrt{5}-1)^2}+\sqrt{(\sqrt{5}+1)^2}`
`=|\sqrt{5}-1|+|\sqrt{5}+1|`
`=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}`
Vậy `\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}=2\sqrt{5}`
$\\$
`b)` `\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}`
`=\sqrt{3^2+2.3.\sqrt{2}+2}-\sqrt{3^2-2.3.\sqrt{2}+2}`
`=\sqrt{(3+\sqrt{2})^2}-\sqrt{(3-\sqrt{2})^2}`
`=|3+\sqrt{2}|-|3-\sqrt{2}|`
`=3+\sqrt{2}-(3-\sqrt{2})`
`=3-3+\sqrt{2}+\sqrt{2}=2\sqrt{2}`
Vậy: `\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=2\sqrt{2}`
$\\$
`c)` `\sqrt{x^2+6x+9}/{x+3}` `\quad (x\ne -3)`
`=\sqrt{(x+3)^2}/{x+3}`
`=|x+3|/{x+3}`
+) Nếu `x+3> 0<=>x> -3`
`=>|x+3|=x+3`
`=>|x+3|/{x+3}={x+3}/{x+3}=1`
+) Nếu `x+3< 0<=>x< -3`
`=>|x+3|=-(x+3)`
`=>|x+3|/{x+3}={-(x+3)}/{x+3}=-1`
Vậy: $\dfrac{\sqrt{x^2+6x+9}}{x+3}=\begin{cases}1\ nếu\ x> -3\\-1\ nếu\ x< -3\end{cases}$
$\\$
`d)` `{x-8\sqrt{x}-9}/{x-1}`
$ĐKXĐ: \begin{cases}x\ge 0\\x-1\ne 0\end{cases}$`<=>`$\begin{cases}x\ge 0\\x\ne 1\end{cases}$
`\qquad {x-8\sqrt{x}-9}/{x-1}`
`={x-9\sqrt{x}+\sqrt{x}-9}/{x-1}`
`={\sqrt{x}(\sqrt{x}-9)+\sqrt{x}-9}/{x-1}`
`={(\sqrt{x}-9)(\sqrt{x}+1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}`
`={\sqrt{x}-9}/{\sqrt{x}-1}`
Vậy: `{x-8\sqrt{x}-9}/{x-1}={\sqrt{x}-9}/{\sqrt{x}-1}` với `x\ge 0; x\ne 1`
