Các dạng toán biến đổi căn thức và cách làm

$\sqrt{{{x}{2}}-10x+25}$ =$\sqrt{{{[x-5]}{2}}}$ =$\left\{ \begin{align}& x-5,[x\ge 5] \\ & x+5,[x0; B>0]

$\sqrt{\frac{A}{B}}=\frac{\sqrt{A}}{\sqrt{B}}=\frac{\sqrt{A}.\sqrt{B}}{{{[\sqrt{B}]}^{2}}}$ =$\frac{1}{B}[\sqrt{AB}]$

VD: $\sqrt{\frac{3}{5}}=\frac{\sqrt{3}}{\sqrt{5}}=\frac{1}{5}\sqrt{15}$

4, $\sqrt{\frac{A}{B}}$ Đk: A.B >0 ,B$\ne $ 0

$\sqrt{\frac{A}{B}}=\sqrt{\frac{A.B}{{{B}^{2}}}}=\frac{1}{\left| B \right|}.\sqrt{AB}$

VD: $\sqrt{\frac{3}{5}}=\sqrt{\frac{3.5}{{{5}^{2}}}}=\frac{1}{5}\sqrt{15}$

$\sqrt{\frac{-3}{-5}}=\sqrt{\frac{[-3].[-5]}{[-5].[-5]}}$ =$\sqrt{\frac{15}{25}}$ =$\frac{1}{5}$$\sqrt{15}$

5,$\sqrt{{{A}^{2}}.B}=\left| A \right|.\sqrt{B}[B\ge 0]$

VD: $\sqrt{27}=\sqrt{{{3}^{2}}.3}=\left| 3 \right|.\sqrt{3}=3\sqrt{3}$

6, $A\sqrt{B}=\sqrt{{{A}^{2}}.B}[A\ge 0;B\ge 0]$

$-3\sqrt{5}=-\sqrt{{{3}^{2}}.5}=-\sqrt{45}$

7,$\frac{A}{\sqrt{B}}=\frac{A\sqrt{B}}{B}[B>0]$

VD: $\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{5}$

8, $\frac{C}{\sqrt{A}-\sqrt{B}}$ [A$\ge 0;B\ge 0;A\ne B$ ]

\=$\frac{C.[\sqrt{A}+\sqrt{B}]}{[\sqrt{A}+\sqrt{B}].[\sqrt{A}-\sqrt{B}]}=\frac{C.[\sqrt{A}+\sqrt{B}]}{A-B}$

9,$\frac{C}{\sqrt{A}+\sqrt{B}}$ [$A\ge 0;B\ge 0;A\ne B$ ]

\=$\frac{C[\sqrt{A}-\sqrt{B}]}{[\sqrt{A}-\sqrt{B}].[\sqrt{A}+\sqrt{B}]}=\frac{C[\sqrt{A}-\sqrt{B}]}{A-B}$

1. bài tập

VD1:

$a,\sqrt{{{\left[ 3x-1 \right]}^{2}}}$ [Vĩnh Phúc 2017]

$b,\sqrt{3}\left[ \sqrt{12}-\sqrt{3} \right]$ [Ninh Bình 2017]

$c,\sqrt{48}-\sqrt{3}$ [Hà Tĩnh 2017]

Giải

a, $\sqrt{{{\left[ 3a-1 \right]}^{2}}}=\left| 3a-1 \right|=\left\{ \begin{align}& 3a-1\,\,khi\,a\ge \frac{1}{3} \\ & 1-3a\,khi\,a