\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

Câu a bạn coi lại đề

b. ĐKXĐ: \(x\ge0;x\ne1\)

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

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

\(\Leftrightarrow5x+1+2\sqrt{3x\left(2x+1\right)}=3x+2\)

\(\Leftrightarrow2\sqrt{6x^2+3x}=1-2x\) (\(x\le\dfrac{1}{2}\) )

\(\Leftrightarrow4\left(6x^2+3x\right)=4x^2-4x+1\)

\(\Leftrightarrow20x^2+16x-1=0\)

\(\Rightarrow x=\dfrac{-4+\sqrt{21}}{10}\)

Bạn xem lại đề câu a.

ĐKXĐ: ....

\(\Leftrightarrow\sqrt{2x^2+5x+2}=1+2\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)+1+4\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow3\left(2x^2+5x-6\right)+4\sqrt{2x^2+5x-6}-7=0\)

Đặt \(\sqrt{2x^2+5x-6}=a\ge0\)

\(3a^2+4a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+5x-6}=1\Leftrightarrow2x^2+5x-7=0\)

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

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

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

=>x-3=1

=>x=4

a: ĐKXĐ: \(x^2-5x-6>=0\)

=>(x-6)(x+1)>=0

=>\(\left[{}\begin{matrix}x>=6\\x< =-1\end{matrix}\right.\)

\(\sqrt{x^2-5x-6}=x-2\)

=>\(\left\{{}\begin{matrix}x-2>=0\\x^2-5x-6=\left(x-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=2\\x^2-5x-6=x^2-4x+4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=6\\-5x-6=-4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=6\\-x=10\end{matrix}\right.\)

=>\(x\in\varnothing\)

b: ĐKXĐ: \(x\in R\)

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

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

=>|x-4|=4-x

=>x-4<=0

=>x<=4

c: ĐKXĐ: \(x^2-2x>=0\)

=>x(x-2)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =0\end{matrix}\right.\)

\(\sqrt{x^2-2x}=2-x\)

=>\(\left\{{}\begin{matrix}x^2-2x=\left(2-x\right)^2\\x< =2\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}2x=4\\x< =2\end{matrix}\right.\Leftrightarrow x=2\left(nhận\right)\)

d: ĐKXĐ: x>=-27/2

\(\sqrt{2x+27}-6=x\)

=>\(\sqrt{2x+27}=x+6\)

=>\(\left\{{}\begin{matrix}x>=-6\\\left(x+6\right)^2=2x+27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\x^2+12x+36-2x-27=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\x^2+10x+9=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-6\\\left(x+9\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-6\\x\in\left\{-9;-1\right\}\end{matrix}\right.\)

=>x=-1

Kết hợp ĐKXĐ, ta được: x=-1

a.

\(\sqrt{x^2-5x-6}=x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-5x-6=\left(x-2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-5x-6=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x=-10\left(ktm\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

b.

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

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

\(\Leftrightarrow\left|x-4\right|=-\left(x-4\right)\)

\(\Leftrightarrow x-4\le0\)

\(\Rightarrow x\le4\)

c.

\(\sqrt{x^2-2x}=2-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}2-x\ge0\\x^2-2x=\left(2-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x^2-2x=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\2x=4\end{matrix}\right.\)

\(\Rightarrow x=2\)

d.

\(\Leftrightarrow\sqrt{2x+27}=x+6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\x+27=\left(x+6\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x+27=x^2+12x+36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x^2+11x+9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{85}}{2}\\x=\dfrac{-11-\sqrt{85}}{2}\left(loại\right)\end{matrix}\right.\)

1. ĐKXĐ: $\xgeq \frac{-6}{5}$

PT \(\Leftrightarrow [\sqrt{2x^2+5x+7}-(x+3)]+[(x+2)-\sqrt{5x+6}]+(x^2-x-2)=0\)

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

\(\Leftrightarrow (x^2-x-2)\left(\frac{1}{\sqrt{2x^2+5x+7}+x+3}+\frac{1}{x+2+\sqrt{5x+6}}+1\right)=0\)

Với $x\geq \frac{-6}{5}$, dễ thấy biểu thức trong ngoặc lớn hơn hơn $0$

Do đó: $x^2-x-2=0$

$\Leftrightarrow (x+1)(x-2)=0$

$\Leftrightarrow x=-1$ hoặc $x=2$ (đều thỏa mãn)

Bài 2: Tham khảo tại đây:

Giải pt \(\sqrt{2x+1} - \sqrt[3]{x+4} = 2x^2 -5x -11\) - Hoc24

Bài 3:

ĐKXĐ: $\frac{19}{3}\geq x\geq -3$

PT \(\Leftrightarrow x^2+2x+9-4\sqrt{x+3}-\sqrt{19-3x}=0\)

\(\Leftrightarrow (x^2+x-2)+\frac{4}{3}[x+5-3\sqrt{x+3}]+\frac{1}{3}[13-x-3\sqrt{19-3x}]=0\)

\(\Leftrightarrow (x^2+x-2)+\frac{4}{3}.\frac{x^2+x-2}{x+5+3\sqrt{x+3}}+\frac{1}{3}.\frac{x^2+x-2}{13-x+3\sqrt{19-3x}}=0\)

\(\Leftrightarrow (x^2+x-2)\left[1+\frac{4}{3}.\frac{1}{x+5+3\sqrt{x+3}}+\frac{1}{3}.\frac{1}{13-x+3\sqrt{19-3x}}\right]=0\)

Dễ thấy biểu thức trong ngoặc vuông $>0$ nên $x^2+x-2=0$

$\Leftrightarrow (x-1)(x+2)=0$
$\Leftrightarrow x=1$ hoặc $x=-2$ (đều thỏa mãn)