a

\(\sqrt{9\left(2-3x\right)^2}=6\\ \Leftrightarrow3\left|2-3x\right|=6\\ \Leftrightarrow\left|2-3x\right|=2\)

Với \(x\le\dfrac{2}{3}\) thì PT trở thành:

\(2-3x=2\\ \Leftrightarrow3x=0\\ \Leftrightarrow x=0\left(nhận\right)\)

Với \(x>\dfrac{2}{3}\) thì PT trở thành:

\(3x-2=2\\ \Leftrightarrow3x=4\\ \Leftrightarrow x=\dfrac{4}{3}\left(nhận\right)\)

b

ĐK: \(x\ge-\dfrac{3}{2}\)

\(\sqrt{4x^2-9}=2\sqrt{2x+3}\\ \Leftrightarrow\sqrt{\left(2x\right)^2-3^2}=2\sqrt{2x+3}\\ \Leftrightarrow\sqrt{2x-3}.\sqrt{2x+3}-2\sqrt{2x+3}=0\\ \Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=0\\\sqrt{2x-3}-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\2x-3=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(nhận\right)\\x=\dfrac{7}{2}\left(nhận\right)\end{matrix}\right.\)

c

ĐK: \(x\ge3\)

\(\sqrt{10\left(x-3\right)}=\sqrt{20}\\ \Leftrightarrow10\left(x-3\right)=20\\ \Leftrightarrow x-3=2\\ \Leftrightarrow x=5\left(nhận\right)\)

d

\(\sqrt{x^2+6x+9}=3x-6\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-6\\ \Leftrightarrow\left|x+3\right|=3x-6\)

Với \(x\ge-3\) thì PT trở thành:

\(x+3=3x-6\\ \Leftrightarrow x+3-3x+6=0\\ \Leftrightarrow-2x+9=0\\ \Leftrightarrow x=\dfrac{9}{2}\left(nhận\right)\)

Với \(x< -3\) thì PT trở thành:

\(-x-3=3x-6\\ \Leftrightarrow-x-3-3x+6=0\\ \Leftrightarrow-2x+3=0\\ \Leftrightarrow x=\dfrac{3}{2}\left(loại\right)\)

Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

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

\(\sqrt{x^2+6x+9}=3x-6\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-6\\ \Leftrightarrow\left|x+3\right|=3x-6\)

Nếu \(\left\{{}\begin{matrix}x+3\ge0\Leftrightarrow x\ge-3\\x+3< 0\Leftrightarrow x< -3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\\-\left(x+3\right)=3x-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-3x=-6-3\\-x-3=3x-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-9\\-x-3x=-6+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\-4x=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\)

a) \(\sqrt{x}-x-0\) (ĐK: \(x\ge0\))

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\1-\sqrt{x}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

b) \(x-\sqrt{2x-9}=6\)

\(\Leftrightarrow\sqrt{2x-9}=x-6\) (ĐK: \(x\ge\dfrac{9}{2}\))

\(\Leftrightarrow2x-9=\left(x-6\right)^2\)

\(\Leftrightarrow2x-9=x^2-12x+36\)

\(\Leftrightarrow x^2-14x+45=0\)

\(\Leftrightarrow x^2-5x-9x+45=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)

c) \(3x-\sqrt{6x-\left(3-2\right)}=0\) (ĐK: \(x\ge\dfrac{1}{6}\))

\(\Leftrightarrow3x-\sqrt{6x-1}=0\)

\(\Leftrightarrow\sqrt{6x-1}=3x\)

\(\Leftrightarrow6x-1=9x^2\)

\(\Leftrightarrow9x^2-6x+1=0\)

\(\Leftrightarrow\left(3x-1\right)^2=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\left(tm\right)\)

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)