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: \(x\ge\dfrac{1}{6}\))

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

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

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

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

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

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

a)

DK: x\(\ge\)-2,x\(\ge\)\(\dfrac{1}{2}\)

=>\(\sqrt{4\left(x+2\right)}-\sqrt{2x-1}+\sqrt{9\left(x+2\right)}=0\)

\(\Leftrightarrow2\sqrt{x+2}-\sqrt{2x-1}+3\sqrt{x+2}=0\)

\(\Leftrightarrow5\sqrt{x+2}-\sqrt{2x-1}=0\)

\(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\)

<=>25x+50=2x-1

=>23x=-51

=>x=\(-\dfrac{51}{23}\)(ko thỏa mãn dk)

=> phương trình vô nghiệm..

b)

ĐKXĐ:\(x\ge1,x\ge-1\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x-1\right)}-3\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-3\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)(nhận)

Vậy S={1;8}

c) ĐKXĐ:

\(x\ge0\)

\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)

\(\Leftrightarrow-11\sqrt{2x}=-11\)

\(\Leftrightarrow\sqrt{2x}=1\)

\(\Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)

Câu a :\(\sqrt{4x+8}-2\sqrt{2x-1}+\sqrt{9x+18}=0\) ( ĐK : \(x\ge\dfrac{1}{2}\) )

\(\Leftrightarrow\sqrt{4x+8}+\sqrt{9x+18}=\sqrt{2x-1}\)

\(\Leftrightarrow2\sqrt{x+2}+3\sqrt{x+2}=\sqrt{2x-1}\)

\(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\)

\(\Leftrightarrow25\left(x+2\right)=2x-1\)

\(\Leftrightarrow25x+50=2x-1\)

\(\Leftrightarrow23x=-51\)

\(\Leftrightarrow x=-\dfrac{51}{23}< -\dfrac{1}{2}\)

Vậy phương trình vô nghiệm .

Câu b :

\(\sqrt{x^2-1}-\sqrt{9\left(x-1\right)}=0\) ( ĐK : \(x\ge1\) )

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-3\sqrt{\left(x-1\right)}=0\)

\(\Leftrightarrow\sqrt{\left(x-1\right)}\left(\sqrt{x+1}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

Vậy \(S=\left\{1;8\right\}\)

Câu c : \(\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5\) ( ĐK : \(x\ge\dfrac{5}{6}\) )

\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)

\(\Leftrightarrow-11\sqrt{2x}+11=0\)

\(\Leftrightarrow-11\left(\sqrt{2x}-1\right)=0\)

\(\Leftrightarrow\sqrt{2x}-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)

Vậy \(S=\left\{\dfrac{1}{2}\right\}\)

Chúc bạn học tốt

a.

ĐKXĐ: \(x\ge3\)

(Tốt nhất bạn kiểm tra lại đề cái căn đầu tiên của \(\sqrt{x-3}\) là căn bậc 2 hay căn bậc 3). Vì nhìn ĐKXĐ thì thấy căn bậc 2 là không hợp lý rồi đó

Pt tương đương:

\(\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)=0\)

Do \(x\ge3\Rightarrow x-2>0\Rightarrow\left(x+1\right)\left(x-2\right)>0\)

\(\Rightarrow\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)>0\)

Pt vô nghiệm

b.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

Pt: \(2x+3-\sqrt{2x+3}-\left(4x^2-6x+2\right)=0\)

Đặt \(\sqrt{2x+3}=t\ge0\) ta được:

\(t^2-t-\left(4x^2-6x+2\right)=0\)

\(\Delta=1+4\left(4x^2-6x+2\right)=\left(4x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t_1=\dfrac{1+4x-3}{2}=2x-1\\t_2=\dfrac{1-4x+3}{2}=2-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=2x-1\left(x\ge\dfrac{1}{2}\right)\\\sqrt{2x+3}=2-2x\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4x^2-4x+1\left(x\ge\dfrac{1}{2}\right)\\2x+3=4x^2-8x+4\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{17}}{4}\\x=\dfrac{5-\sqrt{21}}{4}\end{matrix}\right.\)

c.

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow2\left(x^2-4x+4\right)+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)

\(\Leftrightarrow2\left(x-2\right)^2-5\left(x-2\right)\sqrt{x+1}+2\left(x+1\right)=0\)

Đặt \(\left\{{}\begin{matrix}x-2=a\\\sqrt{x+1}=b\end{matrix}\right.\) ta được:

\(2a^2-5ab+2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=x-2\left(x\ge2\right)\\\sqrt{x+1}=2x-4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-4x+4\\x+1=4x^2-16x+16\end{matrix}\right.\) (\(x\ge2\))

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\) (đã loại nghiệm)

ĐKXĐ:...

a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành:

\(3a^2-2b^2+ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)

\(\Leftrightarrow3a=2b\)

\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

Phương trình trở thành:

\(a^2+2+ab=3a+b\)

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

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

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

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

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

\(\Leftrightarrow...\)

a/ ĐKXĐ: ...

\(\Leftrightarrow\left(x^2-6x\right)\left(\sqrt{17-x^2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x=0\\\sqrt{17-x^2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-6\right)=0\\x^2=16\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\left(l\right)\\x=4\\x=-4\end{matrix}\right.\)

b/ĐKXĐ: \(x\ge-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=0\\\sqrt{x+3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\left(l\right)\\x=-3\end{matrix}\right.\)

c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ge1\\x\le1\end{matrix}\right.\) \(\Rightarrow x=1\)

Thay \(x=1\) vào pt thấy ko thỏa mãn

Vậy pt vô nghiệm

d/ ĐKXĐ: \(x\ge2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\left(l\right)\\x=2\end{matrix}\right.\)

e/ ĐKXĐ: \(x\ge\frac{-3+\sqrt{17}}{2}\)

\(\Leftrightarrow x^2+3x-2=x+1\)

\(\Leftrightarrow x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\left(l\right)\end{matrix}\right.\)

f/ ĐKXĐ: \(x\ge4\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-4}-1=0\\x^2-7x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=1\\\left(x-1\right)\left(x-6\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\left(l\right)\\x=6\end{matrix}\right.\)