ĐKXĐ: \(\left\{{}\begin{matrix}x+8>=0\\6-2x>=0\end{matrix}\right.\Leftrightarrow-8< =x< =3\)

\(PT\Leftrightarrow1=6\sqrt{x+8}-18+x+4\sqrt{6-2x}-8\)

\(\Leftrightarrow6\cdot\dfrac{x+8-9}{\sqrt{x+8}+3}+x-1+4\cdot\dfrac{6-2x-4}{\sqrt{6-2x}+2}=0\)

=>\(\left(x-1\right)\left(\dfrac{6}{\sqrt{x+8}+3}+1-\dfrac{4}{\sqrt{6-2x}+2}\right)=0\)

=>x-1=0

=>x=1

(x² + 5x + 6)(x² + 9x + 20) = 24

<=> (x + 2)(x + 3)(x + 4)(x + 5) - 24 = 0

<=> (x² + 7x + 10)(x² + 7x + 12) - 24 = 0 (1)

Đặt x² + 7x + 11 = t, ta có:

(1) <=> (t - 1)(t + 1) - 24 = 0

<=> t² - 1 - 24 = 0

<=> (t - 5)(t + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}t-5=0\\t+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+7x+11-5=0\\x^2+7x+11+5=0\end{matrix}\right.\)

<=> (x + 1)(x + 6) = 0 (vì \(x^2+7x+16\ge\dfrac{15}{4}>0\))

<=> x = - 1 hoặc x = - 6

~ ~ ~ ~ ~

x⁴ - 24x = 32

<=> x⁴ - 24x - 32 = 0

<=> (x² - 2x - 4)(x² + 2x + 8) = 0

<=> \(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)=0\) (vì \(x^2+2x+8\ge7>0\))

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

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

ĐKXĐ: \(-\frac{3}{2}\le x\le12\)

\(\Leftrightarrow x^2-2x\sqrt{2x+3}+2x+3+12-x-6\sqrt{12-x}+9=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+3}\right)^2+\left(\sqrt{12-x}-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2x+3}=0\\\sqrt{12-x}-3=0\end{matrix}\right.\) \(\Rightarrow x=3\)

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

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

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

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

=>x-6=3-2x hoặc x-6=2x-3

=>3x=9 hoặc -x=3

=>x=3 hoặc x=-3

\(\left|x-6\right|=\left|3-2x\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-6=3-2x\\x-6=2x-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-6-3+2x=0\\2x-3-x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x-9=0\\x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

<=> \(x^2\left(x+3\right)-2\left(x+3\right)=0\)

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