a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

Câu b còn 1 cách giải nữa:

Với \(x=0\) không phải nghiệm

Với \(x>0\) , chia 2 vế cho \(\sqrt{x}\) ta được:

\(\sqrt{2x+8+\dfrac{5}{x}}+\sqrt{2x-4+\dfrac{5}{x}}=6\)

Đặt \(\sqrt{2x-4+\dfrac{5}{x}}=t>0\Leftrightarrow2x+8+\dfrac{5}{x}=t^2+12\)

Phương trình trở thành:

\(\sqrt{t^2+12}+t=6\)

\(\Leftrightarrow\sqrt{t^2+12}=6-t\)

\(\Leftrightarrow\left\{{}\begin{matrix}6-t\ge0\\t^2+12=\left(6-t\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\le6\\12t=24\end{matrix}\right.\)

\(\Rightarrow t=2\)

\(\Rightarrow\sqrt{2x-4+\dfrac{5}{x}}=2\)

\(\Leftrightarrow2x-4+\dfrac{5}{x}=4\)

\(\Rightarrow2x^2-8x+5=0\)

\(\Leftrightarrow...\)

a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

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

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

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

\(\Leftrightarrow x^2-2x-4=0\)

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

c, ĐK: \(0\le x\le9\)

Đặt \(\sqrt{9x-x^2}=t\left(0\le t\le\dfrac{9}{2}\right)\)

\(pt\Leftrightarrow9+2\sqrt{9x-x^2}=-x^2+9x+m\)

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

\(\Leftrightarrow-t^2+2t+9=m\)

Khi \(m=9,pt\Leftrightarrow-t^2+2t=0\Leftrightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9x-x^2=0\\9x-x^2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=9\left(tm\right)\\x=\dfrac{9\pm\sqrt{65}}{2}\left(tm\right)\end{matrix}\right.\)

Phương trình đã cho có nghiệm khi phương trình \(m=f\left(t\right)=-t^2+2t+9\) có nghiệm

\(\Leftrightarrow minf\left(t\right)\le m\le maxf\left(t\right)\)

\(\Leftrightarrow-\dfrac{9}{4}\le m\le10\)

=>\(\dfrac{x^2-3x+6-x^2+3x-3}{\sqrt{x^2-3x+6}-\sqrt{x^2-3x+3}}=3\)

=>căn x^2-3x+6 - căn x^2-3x+3=1

Đặt x^2-3x+3=a

=>căn a+3-căn a=1

=>a+3+a-2căn a(a+3)=1

=>2căn a(a+3)=2a+3-1=2a+2

=>căn a(a+3)=a+1

=>a^2+3a=a^2+2a+1

=>a=1

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

=>x=1 hoặc x=2

=>\(x^2+9-12\sqrt{x^2-25}=13x+5-12\sqrt{x^2-25}\)

<=> \(x^2-13x+4=0\)

........

\(=>x^2+11-12\sqrt{x^2-25}=13x+25-12\sqrt{x^2-25}\)

\(< =>x^2-13x-14=0\)

\(< =>\left(x+1\right)\left(x-14\right)=0\)

..............

a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)

Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)

Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)

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

\(\Leftrightarrow b=a\)

Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)

\(\Leftrightarrow x^3-4x^2-6x+5=0\)

\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)

`sqrt{x^2-25}-6=3sqrt{x+5}-2sqrt{x-5}(x>=5)`

`<=>sqrt{(x-5)(x+5)}+2sqrt{x-5}=3sqrt{x+5}+6`

`<=>sqrt{x-5}(sqrt{x+5}+2)=3(sqrt{x+5}+2)`

`<=>(sqrt{x+5}+2)(sqrt{x-5}-3)=0`

Vì `sqrt{x+5}+2>0`

`<=>sqrt{x-5}-3=0`

`<=>sqrt{x-5}=3`

`<=>x-5=9<=>x=14(tm)`

Vậy `x=14`

\(\sqrt{x^2-25}-6=3\sqrt{x+5}-2\sqrt{x-5}\\ \Leftrightarrow\sqrt{\left(x-5\right)\left(x+5\right)}-6-3\sqrt{x+5}+2\sqrt{x-5}=0\\ \Leftrightarrow\left(2\sqrt{x-5}+\sqrt{\left(x-5\right)\left(x+5\right)}\right)-\left(3\sqrt{x+5}+6\right)=0\Leftrightarrow\sqrt{x-5}\left(2+\sqrt{x+5}\right)-3\left(2+\sqrt{x+5}\right)=0\\ \Leftrightarrow\left(\sqrt{x-5}-3\right)\left(2+\sqrt{x-5}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-5}=3\\\sqrt{x-5}=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=9\\x\in\varnothing\end{matrix}\right.\Leftrightarrow x=14\)

Lời giải:
ĐKXĐ: $\frac{2}{3}\leq x\leq 6$

PT $\Leftrightarrow 3(\sqrt{3x-2}-2)+x(\sqrt{6-x}-2)=2(2-x)$

$\Leftrightarrow (2-x)(2-\frac{x}{\sqrt{6-x}+2}+\frac{9}{\sqrt{3x-2}+2})=0$

Với $\frac{2}{3}\leq x\leq 6$ thì $2+\frac{9}{\sqrt{3x-2}+2}\geq \frac{7}{2}>3$ còn $\frac{x}{\sqrt{6-x}+2}\leq \frac{6}{2}=3$ nên biểu thức $2-\frac{x}{\sqrt{6-x}+2}+\frac{9}{\sqrt{3x-2}+2}>0$

$\Rightarrow 2-x=0$

$\Leftrightarrow x=2$ (tm)

ĐKXĐ: \(x\ge1\)

\(\sqrt[3]{x+6}-2+\sqrt[]{x-1}-1=x^2-4\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{x-2}{\sqrt[]{x-1}+1}=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt[]{x-1}+1}=x+2\left(1\right)\end{matrix}\right.\)

Xét (1); do \(x\ge1\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[2]{x+6}+4}+\dfrac{1}{\sqrt[]{x-1}+1}< \dfrac{1}{4}+1< 2\\x+2>1+2>2\end{matrix}\right.\)

\(\Rightarrow\left(1\right)\) vô nghiệm

Vậy \(x=2\) là nghiệm duy nhất của pt

a: ĐKXĐ: \(\left\{{}\begin{matrix}x-3>=0\\5-x>=0\end{matrix}\right.\)

=>3<=x<=5

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

=>\(\sqrt{x-3}-1+\sqrt{5-x}-1=0\)

=>\(\dfrac{x-3-1}{\sqrt{x-3}+1}+\dfrac{5-x-1}{\sqrt{5-x}+1}=0\)

=>\(\left(x-4\right)\left(\dfrac{1}{\sqrt{x-3}+1}-\dfrac{1}{\sqrt{5-x}+1}\right)=0\)

=>x-4=0

=>x=4