a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

\(\Leftrightarrow\sqrt{x-2}=4\)

=>x-2=16

hay x=18

b: \(\Leftrightarrow\left|3x+2\right|=4x\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)

c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

\(\Leftrightarrow4\sqrt{x-2}=40\)

=>x-2=100

hay x=102

d: =>5x-6=9

hay x=3

\(a,\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\left(dk:x\ge2\right)\)

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

\(\Leftrightarrow\sqrt{x-2}=4\)

\(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\left(tmdk\right)\)

b,\(\sqrt{9x^2-12x+4=3x\left(dk:x\ge0\right)}\)

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=\dfrac{1}{3}\left(tmdk\right)\end{matrix}\right.\)

Các câu còn lại làm tương tự nhé 

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)

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

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)

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

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

\(\left|x-2\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

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

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

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

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

=>\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

cái j sao khó nhìn vậy

d,   \(x^8+x^7+1\)

\(=x^8-x^2+x^7-x+x^2+x+1\) 

\(=x^2\left(x^6-1\right)+x\left(x^6-1\right)+x^2+x+1\)

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

\(=\left(x^5+x^2\right)\left(x^3-1\right)+\left(x^4+x\right)\left(x^3-1\right)+x^2+x+1\)

\(=\left(x^3-1\right)\left(x^5+x^4+x^2+x\right)+x^2+x+1\)

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

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x\right)+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

c,    \(x^4+5x^3-12x^2+5x+1\)

\(=x^4-x^3+6x^3-6x^2-6x^2+6x-x+1\)

\(=x^3\left(x-1\right)+6x^2\left(x-1\right)-6x\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left[x^3+6x^2-6x-1\right]\)

\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+6x\left(x-1\right)\right]\)

\(=\left(x-1\right)\left(x-1\right)\left(x^2+7x+1\right)\)

\(=\left(x-1\right)^2.\left(x^2+7x+1\right)\)

a,   \(\left(x^2+x-2\right)\left(x^2+9x+18\right)-28\)

\(=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28\)

\(=\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]-28\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-28\)

\(=\left(x^2+5x\right)^2-36-28\)

\(=\left(x^2+5x\right)^2-64\)

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

b, \(B=\left(x+1\right)^2\left(2x+3\right)-18\)

       \(=\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18\)

Đặt \(x^2+2x+1=t\Rightarrow4x^2+8x+3=4t-1\)

Ta có: \(B=\left(4t-1\right)t-18\)

             \(=4t^2-t-18\)

             \(=4t^2-9t+8t-18\)

             \(=t\left(4t-9\right)+2\left(4t-9\right)\)

             \(=\left(4t-9\right)\left(t+2\right)\)

             \(=\left(4x^2+8x-5\right)\left(x^2+2x+3\right)\) (vì \(t=x^2+2x+1\)

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

Chúc bạn học tốt.

a: \(=\dfrac{x^5}{3x}+\dfrac{12x^2}{3x}+\dfrac{5x}{3x}=\dfrac{1}{3}x^4+4x+\dfrac{5}{3}\)

b: \(=\dfrac{-5x^4}{-5x}-\dfrac{15x^3}{5x}+\dfrac{18x}{5x}=x^3-3x^2+\dfrac{18}{5}\)

c: \(=\dfrac{-x^6}{0.5x}+\dfrac{5x^4}{0.5x}-\dfrac{2x^3}{0.5x}=-2x^5+10x^3-4x^2\)

Với \(x=0\) không phải nghiệm

Với \(x\ne0\) chia 2 vế cho \(x^2\) ta được:

\(x^2-5x-12-\dfrac{5}{x}+\dfrac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}+2\right)-5\left(x+\dfrac{1}{x}\right)-14=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-14=0\)

Đặt \(x+\dfrac{1}{x}=t\)

\(\Rightarrow t^2-5t-14=0\Rightarrow\left[{}\begin{matrix}t=7\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=-2\\x+\dfrac{1}{x}=7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\x^2-7x+1=0\end{matrix}\right.\) (bấm máy)

a/ Chắc là bạn ghi nhầm đề? Số cuối là số 9 mới đúng, chứ 27 thì câu này vô nghiệm

\(x^4+4x^3+4x^2+8x^2+12x+27=0\)

\(\Leftrightarrow x^2\left(x+2\right)^2+8\left(x+\frac{3}{4}\right)^2+\frac{45}{2}=0\)

Vế phải dương nên pt vô nghiệm

b/ Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\) ta được:

\(x^2+\frac{1}{x^2}-5\left(x-\frac{1}{x}\right)+6=0\)

Đặt \(x-\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2+2\)

\(\Rightarrow a^2+2-5a+6=0\)

\(\Leftrightarrow a^2-5a+8=0\Rightarrow\) pt vô nghiệm

Lại nhầm đề nữa???? Dấu thứ 2 là dấu + thì pt này có nghiệm đẹp