# Bài 1: Căn bậc hai

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Bài 1: Giải PT

a) $\sqrt{x^2-1}-x^2+1=0$

b) $\sqrt{x^2-4}-x+2=0$

c) $\sqrt{x^4-8x^2+16}=2-x$

d) $\sqrt{9x^2+6x+1}\sqrt{11-6\sqrt{2}}$

e) $\sqrt{4^2-9}=2\sqrt{2x+3}$

f) $\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4$

3 tháng 8 2018 lúc 20:20

a) Đk: $\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.$

$\sqrt{x^2-1}-x^2+1=0$

$\Leftrightarrow x^2-1-\sqrt{x^2-1}= 0$

$\Leftrightarrow\left(\sqrt{x^2-1}-1\right)\sqrt{x^2-1}=0$

$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}-1=0\\\sqrt{x^2-1}=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=1\\x^2-1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x^2=2\left(1\right)\\x^2=1\left(2\right)\end{matrix}\right.$

$\left(1\right)\Leftrightarrow x=\pm\sqrt{2}\left(N\right)$

$\left(2\right)\Leftrightarrow x=\pm1\left(N\right)$

Kl: $x=\pm\sqrt{2}$, $x=\pm1$

b) Đk: $\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.$

$\sqrt{x^2-4}-x+2=0$

$\Leftrightarrow\sqrt{x^2-4}=x-2$

$\Leftrightarrow\left\{{}\begin{matrix}x^2-4=x^2-4x+4\\x\ge2\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}4x=8\\x\ge2\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}x=2\left(N\right)\\x\ge2\end{matrix}\right.$

kl: x=2

c) $\sqrt{x^4-8x^2+16}=2-x$

$\Leftrightarrow\sqrt{\left(x^2-4\right)^2}=2-x$

$\Leftrightarrow\left|x^2-4\right|=2-x$ (*)

Th1: $x^2-4< 0\Leftrightarrow-2< x< 2$

(*) $\Leftrightarrow x^2-4=x-2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=-1\left(N\right)\end{matrix}\right.$

Th2: $x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.$

(*)$\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.$

Kl: x=-3, x=-1,x=2

d) $\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}$

$\Leftrightarrow\sqrt{\left(3x+1\right)^2}=\sqrt{\left(3-\sqrt{2}\right)^2}$

$\Leftrightarrow\left|3x+1\right|=3-\sqrt{2}$ (*)

Th1: $3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}$

(*) $\Leftrightarrow3x+1=3-\sqrt{2}\Leftrightarrow x=\dfrac{2-\sqrt{2}}{3}\left(N\right)$

Th2: $3x+1< 0\Leftrightarrow x< -\dfrac{1}{3}$

(*) $\Leftrightarrow3x+1=-3+\sqrt{2}\Leftrightarrow x=\dfrac{-4+\sqrt{2}}{3}\left(N\right)$

Kl: $x=\dfrac{2-\sqrt{2}}{3}$, $x=\dfrac{-4+\sqrt{2}}{3}$

e) Đk: $x\ge-\dfrac{3}{2}$

$\sqrt{4^2-9}=2\sqrt{2x+3}$ $\Leftrightarrow\sqrt{7}=2\sqrt{2x+3}$ $\Leftrightarrow7=8x+12$

$\Leftrightarrow8x=-5\Leftrightarrow x=-\dfrac{5}{8}\left(N\right)$

kl: $x=-\dfrac{5}{8}$

f) Đk: x >/ 5

$\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4$

$\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow2\sqrt{x-5}=4$

$\Leftrightarrow\sqrt{x-5}=2$

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9\left(N\right)$

kl: x=9

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