Question

[Nguyễn Kim Chi]

Giải các phương trình vô tỉ (Phương trình có chứa căn thức)

1) \(\sqrt{x^2-20x+100}=10\)

2) \(\sqrt{x+2\sqrt{x}+1}=6\)

3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)

5) \(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}\)

6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)

7) \(\sqrt{2x^2-2x\sqrt{6}+3}-\sqrt{5-\sqrt{24}}=0\)

8) \(\sqrt{3-2\sqrt{2}}-\sqrt{x^2-2x\sqrt{2}+2}=0\)

9) \(\sqrt{11-\sqrt{120}}=\sqrt{5x^2+x\sqrt{120}+6}\)

Answer 1

[Cô Nàng Song Tử]

1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)

<=> \(\sqrt{\left(x-10\right)^2}=10\)

<=> \(\left|x-10\right|=10\)

=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)

Vậy S = \(\left\{20;0\right\}\)

2) \(\sqrt{x +2\sqrt{x}+1}=6\)

<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)

<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)

<=> \(\left|\sqrt{x}+1\right|=6\)

=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)

Vậy S = \(\left\{25\right\}\)

3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

<=> \(\sqrt{\left(x-3\right)^2}=\sqrt{\sqrt{3^2}+2.\sqrt{3}.1+1^2}\)

<=> \(\left|x-3\right|=\sqrt{\left(\sqrt{3}+1\right)^2}\)

<=> \(\left|x-3\right|=\sqrt{3}+1\)

=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)

Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)

Answer 2

[Cô Nàng Song Tử]

4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)

<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)

<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)

<=> \(\left|\sqrt{3x}+1\right|=5\)

=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)

5) \(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}\)

<=> \(\sqrt{\left(x-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)

<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)

Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)

6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)

<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)

<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)

<=> \(\left|\sqrt{6x}+2\right|=7\)

=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)

=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)