a: ĐKXĐ: 2<=x<=10

\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)

=>\(\sqrt{x-2}-2+\sqrt{10-x}-2=x^2-12x+36\)

=>\(\frac{x-2-4}{\sqrt{x-2}+2}+\frac{10-x-4}{\sqrt{10-x}+2}=\left(x-6\right)^2\)

=>\(\left(x-6\right)\cdot\left(\frac{1}{\sqrt{x-2}+2}-\frac{1}{\sqrt{10-x}+2}-x+6\right)=0\)

=>x-6=0

=>x=6(nhận)

b: ĐKXĐ: 5/3<=x<=7/3

\(\sqrt{3x-5}+\sqrt{7-3x}=5x^2-20x+22\)

=>\(\sqrt{3x-5}-1+\sqrt{7-3x}-1=5x^2-20x+20\)

=>\(\frac{3x-5-1}{\sqrt{3x-5}+1}+\frac{7-3x-1}{\sqrt{7-3x}+1}=5\left(x^2-4x+4\right)\)

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

=>(x-2)\(\left\lbrack\frac{3}{\sqrt{3x-5}+1}-\frac{3}{\sqrt{7-3x}+1}-5\left(x-2\right)\right\rbrack=0\)

=>x-2=0

=>x=2(nhận)

c: ĐKXĐ: x∈R

\(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)

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

=>|x-2|+|x-3|=1

mà \(\left|x-2\right|+\left|x-3\right|\ge\left|x-2-x+3\right|=1\forall x\)

nên dấu '=' xảy ra khi (x-2)(x-3)<=0

=>2<=x<=3

a.\(2\sqrt{12x}-3\sqrt{3x}+4\sqrt{48x}=17\)

=>\(4\sqrt{3x}-3\sqrt{3x}+16\sqrt{3x}=17\)

=>\(17\sqrt{3x}=17\)

=>\(\sqrt{3x}=1\)

=>\(x=\dfrac{1}{3}\)

b.Ta có:\(\sqrt{x^2-6x+9}=1\)

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

=>\(\left|x-3\right|=1\)

Vậy có hai trường hợp:

TH1:\(x-3=1\)

=>\(x=4\)

TH2:\(x-3=-1\)

=>\(x=2\)

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: \(\left(x^2-3x+2\right)\left(x^2-12x+32\right)\le4x^2\)

=>(x-1)(x-2)(x-4)(x-8)<=\(4x^2\)

=>\(\left(x^2-9x+8\right)\left(x^2-6x+8\right)\le4x^2\)

=>\(\left(x^2+8\right)^2-15x\left(x^2+8\right)+54x^2-4x^2\le0\)

=>\(\left(x^2+8\right)^2-15x\left(x^2+8\right)+50x^2\le0\)

=>\(\left(x^2-5x+8\right)\left(x^2-10x+8\right)\le0\)

mà \(x^2-5x+8=x^2-5x+\frac{25}{4}+\frac74=\left(x-\frac52\right)^2+\frac74>0\forall x\)

nên \(x^2-10x+8\le0\)

=>\(x^2-10x+25-17\le0\)

=>\(\left(x-5\right)^2\le17\)

=>\(-\sqrt{17}\le x-5\le\sqrt{17}\)

=>\(-\sqrt{17}+5\) <=x<=\(\sqrt{17}+5\)