Lời giải:

ĐK:.......

Đặt $4x^2+4x+5=a\Rightarrow 8x^2+8x+11=2a+1; 4-4x^2-4x=9-a$

PT trở thành:

$\sqrt{a}+\sqrt{2a+1}=9-a\Leftrightarrow \sqrt{a}-2+\sqrt{2a+1}-3+(a-4)=0$

$\Leftrightarrow \frac{a-4}{\sqrt{a}+2}+\frac{2(a-4)}{\sqrt{2a+1}+3}+(a-4)=0$

$\Leftrightarrow (a-4)\left(\frac{1}{\sqrt{a}+2}+\frac{2}{\sqrt{2a+1}+3}+1\right)=0$

Dễ thấy biểu thức trong ngoặc lớn luôn lớn hơn $0$ nên $a-4=0$

$\Rightarrow a=4$

$\Leftrightarrow 4x^2+4x+5=4$

$\Leftrightarrow 4x^2+4x+1=0\Leftrightarrow (2x+1)^2=0\Leftrightarrow x=-\frac{1}{2}$

1/ ĐKXĐ: $4x^2-4x-11\geq 0$

PT $\Leftrightarrow \sqrt{4x^2-4x-11}=2(4x^2-4x-11)-6$

$\Leftrightarrow a=2a^2-6$ (đặt $\sqrt{4x^2-4x-11}=a, a\geq 0$)

$\Leftrightarrow 2a^2-a-6=0$

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

Vì $a\geq 0$ nên $a=2$

$\Leftrightarrow \sqrt{4x^2-4x-11}=2$

$\Leftrightarrow 4x^2-4x-11=4$

$\Leftrightarrow 4x^2-4x-15=0$
$\Leftrightarrow (2x-5)(2x+3)=0$

$\Rightarrow x=\frac{5}{2}$ hoặc $x=\frac{-3}{2}$ (tm)

2/ ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{3x^2+9x+8}=\frac{1}{3}(3x^2+9x+8)-\frac{14}{3}$

$\Leftrightarrow a=\frac{1}{3}a^2-\frac{14}{3}$ (đặt $\sqrt{3x^2+9x+8}=a, a\geq 0$)

$\Leftrightarrow a^2-3a-14=0$

$\Rightarrow a=\frac{3+\sqrt{65}}{2}$ (do $a\geq 0$)

$\Leftrightarrow 3x^2+9x+8=\frac{37+3\sqrt{65}}{2}$

$\Rightarrow x=\frac{1}{2}(-3\pm \sqrt{23+2\sqrt{65}})$

3. ĐKXĐ: $x^2+3x\geq 0$

PT $\Leftrightarrow 10-(x^2+3x)=\sqrt{x^2+3x}$

$\Leftrightarrow 10-a^2=a$ (đặt $\sqrt{x^2+3x}=a, a\geq 0$)

$\Leftrightarrow a^2+a-10=0$

$\Rightarrow a=\frac{-1+\sqrt{41}}{2}$

$\Leftrightarrow x^2+3x=a^2=\frac{21-\sqrt{41}}{2}$

$\Rightarrow x=\frac{1}{2}(-3\pm \sqrt{51-2\sqrt{41}})$ (đều tm)

a:

ĐKXĐ: \(x>=-2\)

\(1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)

=>\(1+\sqrt{\left(x+2\right)\left(x+5\right)}=\sqrt{x+5}+\sqrt{x+2}\)

Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\)(ĐK: a>0 và b>0)

Phương trình sẽ trở thành:

1+ab=a+b

=>ab-a-b+1=0

=>a(b-1)-(b-1)=0

=>(b-1)(a-1)=0

=>\(\left\{{}\begin{matrix}a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)

=>\(\left\{{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\)

=>\(x\in\varnothing\)

b: \(\sqrt{4x^2-2x+\dfrac{1}{4}}=4x^3-x^2+8x-2\)

=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)

=>\(\sqrt{\left(2x-\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)

=>\(\left|2x-\dfrac{1}{2}\right|=4x^3-x^2+8x-2\)(1)

TH1: x>=1/4

\(\left(1\right)\Leftrightarrow4x^3-x^2+8x-2=2x-\dfrac{1}{2}\)

=>\(4x^3-x^2+6x-\dfrac{3}{2}=0\)

=>\(x^2\left(4x-1\right)+1,5\left(4x-1\right)=0\)

=>\(\left(4x-1\right)\left(x^2+1,5\right)=0\)

=>4x-1=0

=>x=1/4(nhận)

TH2: x<1/4

Phương trình (1) sẽ trở thành:

\(4x^3-x^2+8x-2=-2x+\dfrac{1}{2}\)

=>\(x^2\left(4x-1\right)+2\left(4x-1\right)+0,5\left(4x-1\right)=0\)

=>\(\left(4x-1\right)\cdot\left(x^2+2,5\right)=0\)

=>4x-1=0

=>x=1/4(loại)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

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

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

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

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

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

a/ ĐKXĐ: ....

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

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

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

\(\Leftrightarrow2a^2+2b^2=5ab\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+10=0\\4x^2+3x+10=0\end{matrix}\right.\)

Phương trình vô nghiệm

b/ ĐKXĐ: ....

\(\Leftrightarrow2x^2-x+1=\sqrt{4x^4+4x^2+1-4x^2}\)

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

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

\(\Leftrightarrow\frac{3}{4}\left(2x^2-2x+1\right)+\frac{1}{4}\left(2x^2+2x+1\right)=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)

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

\(\Leftrightarrow3a^2+b^2=4ab\Leftrightarrow3a^2-4ab+b^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x+1=2x^2+2x+1\\9\left(2x^2-2x+1\right)=2x^2+2x+1\end{matrix}\right.\)

c/

\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2\right)^2+16x^2+1-16x^2}\)

\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2+1\right)^2-16x^2}\)

\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)}\)

\(\Leftrightarrow3\left(8x^2+4x+1\right)-2\left(8x^2-4x+1\right)=\sqrt{\left(8x^2+4x+1\right)\left(8x^2-4x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{8x^2+4x+1}=a>0\\\sqrt{8x^2-4x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow3a^2-2b^2=ab\Leftrightarrow3a^2-ab-2b^2=0\)

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

\(\Leftrightarrow a=b\Leftrightarrow\sqrt{8x^2+4x+1}=\sqrt{8x^2-4x+1}\)

\(\Leftrightarrow8x^2+4x+1=8x^2-4x+1\)

\(\Leftrightarrow x=0\)

a) \(4\sqrt{2x+1}-\sqrt{8x+4}+\dfrac{1}{2}\sqrt{32x+16}=12\) (ĐK: \(x\ge-\dfrac{1}{2}\)) 

\(\Leftrightarrow4\sqrt{2x+1}-\sqrt{4\left(2x+1\right)}+\dfrac{1}{2}\cdot4\sqrt{2x+1}=12\)

\(\Leftrightarrow4\sqrt{2x+1}-2\sqrt{2x+1}+2\sqrt{2x+1}=12\)

\(\Leftrightarrow4\sqrt{2x+1}=12\)

\(\Leftrightarrow\sqrt{2x+1}=\dfrac{12}{4}\)

\(\Leftrightarrow2x+1=3^2\)

\(\Leftrightarrow2x=9-1\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=\dfrac{8}{2}\)

\(\Leftrightarrow x=4\left(tm\right)\)

b) \(\sqrt{4x^2-4x+1}=5\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)

\(\Leftrightarrow\left|2x-1\right|=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

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

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

c) \(\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}=-\dfrac{1}{2}\)(ĐK: \(x\ge0;x\ne1\))

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

\(\Leftrightarrow-\sqrt{x}+1=4\sqrt{x}-6\)

\(\Leftrightarrow4\sqrt{x}+\sqrt{x}=1+6\)

\(\Leftrightarrow5\sqrt{x}=7\)

\(\Leftrightarrow\sqrt{x}=\dfrac{7}{5}\)

\(\Leftrightarrow x=\dfrac{49}{25}\left(tm\right)\)