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

1: Ta có: \(\sqrt{4x^2-12x+9}=3-2x\)

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

\(\Leftrightarrow\left(2x-3\right)^2-\left(3-2x\right)^2=0\)

\(\Leftrightarrow\left[\left(2x-3\right)-\left(3-2x\right)\right]\left[\left(2x-3\right)+\left(3-2x\right)\right]=0\)

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

\(\Leftrightarrow\left(4x-6\right)\cdot0=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

2) Ta có: \(\sqrt{x^2-2\cdot\sqrt{2}\cdot x+2}=\sqrt{9-4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{8-2\cdot2\sqrt{2}\cdot1+1}-\sqrt{1+2\cdot1\cdot\sqrt{2}+2}\)

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

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

\(\Leftrightarrow\left|x-\sqrt{2}\right|=\sqrt{2}-2\)(*)

Trường hợp 1: \(x\ge\sqrt{2}\)

(*)\(\Leftrightarrow x-\sqrt{2}=\sqrt{2}-2\)

\(\Leftrightarrow x-\sqrt{2}-\sqrt{2}+2=0\)

\(\Leftrightarrow x-2\sqrt{2}+2=0\)

\(\Leftrightarrow x=2\sqrt{2}-2\)(loại)

Trường hợp 2: \(x< \sqrt{2}\)

(*)\(\Leftrightarrow\sqrt{2}-x=\sqrt{2}-2\)

\(\Leftrightarrow\sqrt{2}-x-\sqrt{2}+2=0\)

\(\Leftrightarrow2-x=0\)

hay x=2(loại)

Vậy: S=∅

\(1.4x^2-12x+9=9-12x+4x^2\)

\(0x=0\)

Pt tm với mọi x

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

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

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

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

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

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

=>\(-2\sqrt{3x}=-4\)

=>\(\sqrt{3x}=2\)

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)

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

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

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

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

1) chả biết nên làm thế nào nữa, đinh chỉ xét dấu thôi là xong, nhưng đang ám ảnh bài giống giống này bị sai

2) Tìm đkxđ --> bình phương 2 vế --> bấm máy tính giải pt bậc 2 --> kl

3) giống câu 2

4) + ĐK: \(x^2-8x+16\ge0\)

pt đã cho \(\Leftrightarrow\left[{}\begin{matrix}\left|x-4\right|+\left|x+2\right|=0\\\left|4-x\right|+\left|x+2\right|=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\end{matrix}\right.\) (vô lý)

Kl: ptvn

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3