a.

ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1-x^2}=t\Rightarrow0\le t\le1\)

\(x^2=1-t^2\Rightarrow x^4=t^4-2t^2+1\)

Pt trở thành:

\(729\left(t^4-2t^2+1\right)+8t=36\)

\(\Leftrightarrow729t^4-1458t^2+8t+693=0\)

\(\Leftrightarrow\left(9t^2+2t-9\right)\left(81t^2-18t-77\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9t^2+2t-9=0\\81t^2-18t-77=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{82}-1}{9}\\t=\dfrac{1+\sqrt{78}}{9}\end{matrix}\right.\)

\(\Rightarrow x=\pm\sqrt{1-t^2}=...\)

b.

ĐKXĐ: ...

\(-3\left(10+4x-x^2\right)-5\sqrt{10+4x-x^2}+42=0\)

Đặt \(\sqrt{10+4x-x^2}=t\ge0\)

\(\Rightarrow-3t^2-5t+42=0\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-\dfrac{14}{3}\left(loại\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow x=...\)

b) \(1+4x-3|x+2|+4=0\)

\(\Leftrightarrow4x-3|x+2|=-5\left(1\right)\)

TH1: Với \(|x+2|=x+2\)thay vào (1) ta được:

\(4x-3\left(x+2\right)=-5\)

\(\Leftrightarrow4x-3x-6=-5\)

\(\Leftrightarrow x=1\)(chọn tự thử lại nhé nó =0 )

TH2: Với \(|x+2|=-x-2\)thay vào (1) ta được: 

\(4x-3\left(-x-2\right)=-5\)

\(\Leftrightarrow4x+3x+6=-5\)

\(\Leftrightarrow7x=-11\)

\(\Leftrightarrow x=\frac{-11}{7}\)( loại tự thử lại nhé nó ko =0 )

Vậy x=1

c.

ĐLXĐ: \(x\ge-\dfrac{1}{3}\)

\(-\left(3x+1\right)+\sqrt{3x+1}+4x^2-10x+6=0\)

Đặt \(\sqrt{3x+1}=t\ge0\)

\(\Rightarrow-t^2+t+4x^2-10x+6=0\)

\(\Delta=1+4\left(4x^2-10x+6\right)=\left(4x-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+4x-5}{-2}=3-2x\\t=\dfrac{-1-4x+5}{-2}=2x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=3-2x\left(x\le\dfrac{3}{2}\right)\\\sqrt{3x-1}=2x-2\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=4x^2-12x+9\left(x\le\dfrac{3}{2}\right)\\3x-1=4x^2-8x+4\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{61}{12}\)

\(\Leftrightarrow36x^2+12x-58-2\sqrt{12x+61}=0\)

\(\Leftrightarrow\left(36x^2+24x+4\right)-\left(12x+61+2\sqrt{12x+61}+1\right)=0\)

\(\Leftrightarrow\left(6x+2\right)^2-\left(\sqrt{12x+61}+1\right)^2=0\)

\(\Leftrightarrow\left(6x+1-\sqrt{12x+61}\right)\left(6x+3+\sqrt{12x+61}\right)=0\)

\(\Leftrightarrow...\) tương tự câu a

a.

ĐKXĐ: \(x\ge-\dfrac{5}{4}\)

\(\Leftrightarrow4x^2-12x-2-2\sqrt{4x+5}=0\)

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

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

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

\(\Leftrightarrow\left(2x-3-\sqrt{4x+5}\right)\left(2x-1+\sqrt{4x+5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+5}=2x-3\left(x\ge\dfrac{3}{2}\right)\\\sqrt{4x+5}=1-2x\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=4x^2-12x+9\left(x\ge\dfrac{3}{2}\right)\\4x+5=4x^2-4x+1\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\sqrt{4x^2-12x+9}+3=2x\)

<=>\(\sqrt{4x^2-12x+9}=2x-3\)

<=>\(4x^2-12x+9=\left(2x-3\right)^2\)

<=>\(4x^2-12x+9=4x^2-12x+9\)

<=>\(4x^2-12x+9-4x^2+12x-9=0\)

<=>0=0( luôn đúng )

=> phương trình trên có vô số nghiệm

Vậy phương trình trên có vô số nghiệm

Ta có: \(\sqrt{4x^2-12x+9}+3=2x\)

\(\Leftrightarrow\left|2x-3\right|=2x-3\)

\(\Leftrightarrow2x-3\ge0\)

hay \(x\ge\dfrac{3}{2}\)

Đặt \(\dfrac{x}{\sqrt{4x-1}}=a\)

Theo đề, ta có phương trình:

a+1/a=2

\(\Leftrightarrow a+\dfrac{1}{a}=2\)

\(\Leftrightarrow\dfrac{a^2+1-2a}{a}=0\)

=>a=1

=>\(x=\sqrt{4x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4x-1\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=3\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow x\in\left\{2+\sqrt{3};2-\sqrt{3}\right\}\)

\(đk:4x^2-12x+44\ge0\left(luôn-đúng\right)\)

\(x^2+\sqrt{4x^2-12x+44}=3x+4\)

\(\Leftrightarrow x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

\(\Leftrightarrow x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

\(đặt:\sqrt{x^2-3x+11}=t\left(t\ge0\right)\)

\(\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-5\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-3x+11}=3\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

ĐKXĐ: ...

\(x^2+\sqrt{4x^2-12x+44}=3x+4\)

\(\Leftrightarrow\sqrt{4x^2-12x+44}=3x+4-x^2\)

\(\Leftrightarrow4x^2-12x+44=\left(3x+4-x^2\right)^2\)

\(\Leftrightarrow4x^2-12x+44=x^4-6x^3+x^2+24x+16\)

\(\Leftrightarrow x^4-6x^3-3x^2+36x-28=0\)

...........