6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

a/ ĐKXĐ: \(0\le x\le4\)

\(\left(x^2-4x\right)\sqrt{-x^2+4x}+x^2-4x+2=0\)

Đặt \(\sqrt{-x^2+4x}=a\ge0\)

\(-a^2.a-a^2+2=0\)

\(\Leftrightarrow a^3+a^2-2=0\)

\(\Leftrightarrow\left(a-1\right)\left(a^2+2a+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a^2+2a+2=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-x^2+4x}=1\Leftrightarrow x^2-4x+1=0\Rightarrow...\)

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

Đặt \(x\sqrt{2x^2+4}=a\Rightarrow x^2\left(2x^2+4\right)=a^2\Rightarrow x^4+2x^2=\frac{a^2}{2}\)

\(\frac{a^2}{2}+a-4=0\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=2\left(x>0\right)\\x\sqrt{2x^2+4}=-4\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^4+4x^2=4\\2x^4+4x^2=16\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=\sqrt{3}-1\\x^2=-\sqrt{3}-1\left(l\right)\\x^2=2\\x^2=-4\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)

c/ Đặt \(\sqrt[3]{2x^2+3x-10}=a\Rightarrow2x^2+3x=a^3+10\)

\(a^3+10-14=2a\)

\(\Leftrightarrow a^3-2a-4=0\)

\(\Leftrightarrow\left(a-2\right)\left(a^2+2a+2\right)=0\Rightarrow a=2\)

\(\Rightarrow\sqrt[3]{2x^2+3x-10}=2\Rightarrow2x^2+3x-18=0\Rightarrow...\)

d/ \(\Leftrightarrow2\left(3x^2+x+4\right)+\sqrt[3]{3x^2+x+4}-18=0\)

Đặt \(\sqrt[3]{3x^2+x+4}=a\)

\(2a^3+a-18=0\)

\(\Leftrightarrow\left(a-2\right)\left(2a^2+4a+9\right)=0\Rightarrow a=2\)

\(\Rightarrow\sqrt[3]{3x^2+x+4}=2\Rightarrow3x^2+x-4=0\Rightarrow...\)

e/ \(\Leftrightarrow x^2+5x+2-3\sqrt{x^2+5x+2}-2=0\)

Đặt \(\sqrt{x^2+5x+2}=a\ge0\)

\(a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=\frac{3+\sqrt{17}}{2}\\a=\frac{3-\sqrt{17}}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+5x+2}=\frac{3+\sqrt{17}}{2}\Rightarrow x^2+5x-\frac{9+3\sqrt{17}}{2}=0\)

Bài cuối xấu quá, chắc nhầm số liệu

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

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

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

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

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

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

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

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

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

<=> \(\sqrt{\left(x\sqrt{2}\right)^2-2x\sqrt{2}.\sqrt{3}+\sqrt{3}^2}=\sqrt{5-\sqrt{4}.\sqrt{6}}\)

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

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

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

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

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

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

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

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

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

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

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

=> \(\left[{}\begin{matrix}x=2\sqrt{2}-1\\x=1\end{matrix}\right.\) Vậy S = \(\left\{2\sqrt{2}-1;1\right\}\)

a: \(\Leftrightarrow6x^2+2x+8+\sqrt{3x^2+x+4}-18=0\)

\(\Leftrightarrow2\left(\sqrt[3]{3x^2+x+4}\right)^3+\sqrt[3]{3x^2+x+4}-18=0\)

=>\(3x^2+x+4=8\)

=>3x^2+x-4=0

=>x=1 hoặc x=-4/3

b: ĐKXĐ: x>0

Pt sẽ là \(x+8+9x-6\sqrt{x\left(x+8\right)}=0\)

=>\(10x+8=\sqrt{36x\left(x+8\right)}\)

=>36x^2+288x=100x^2+160x+64

=>x=1