\(\left(x-2\right)^8=\left(x-2\right)^6\)

\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)

\(\Leftrightarrow\left(x-2\right)^6\left[\left(x-2\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-2\right)^6\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow x=2;x=3;x=1\)

=>x-2=0 hoặc x-2=1

=>x-2=0=>x=2

=>x-2=1=>x=3

\(\left(x-2\right)^8=\left(x-2\right)^6\)

\(\left(x-2\right)^8-\left(x-2\right)^6=0\)

\(\left(x-2\right)^6.\left[\left(x-2\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^6=0\\\left(x-2\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x-2=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

a) Ta có: \(\dfrac{3a^2-10a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a^2-9a-a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a\left(a-3\right)-\left(a-3\right)}{2\left(a-3\right)}\)

\(=\dfrac{\left(a-3\right)\left(3a-1\right)}{2\left(a-3\right)}\)

\(=\dfrac{3a-1}{2}\)

\(=\dfrac{3}{2}a-\dfrac{1}{2}\)(đpcm)

b) Ta có: \(\dfrac{b^2+3b+9}{b^3-27}\)\(=\dfrac{b^2+3b+9}{\left(b-3\right)\left(b^2+3b+9\right)}\)

\(=\dfrac{1}{b-3}\)

\(=\dfrac{b-2}{\left(b-3\right)\left(b-2\right)}\)

\(=\dfrac{b-2}{b^2-5b+6}\)(đpcm)

Help me . Mai mik phải nạp rồi. 

Ai làm đúng mik sẽ

Ai làm đúng mik sẽ k nhé. Thanks

=>x/27=1/9+3/27=1/9+1/9=2/9

=>x=6

`x/27+(-3/27)=1/9`

`=>x/27+(-1/9)=1/9`

`=>x/27=1/9-(-1/9)`

`=>x/27=1/9+1/9`

`=>x/27=2/9`

`=>x/27=6/27`

`=>x=6`

Vậy `x=6`

\(\Leftrightarrow\dfrac{3}{x-2}>0\)

=>x-2>0

hay x>2

\(x+\dfrac{1}{5}-\dfrac{3}{7}=\dfrac{6}{35}\)
\(x+\dfrac{1}{5}=\dfrac{6}{35}+\dfrac{3}{7}\)
\(x+\dfrac{1}{5}=\dfrac{6}{35}+\dfrac{15}{35}\)
\(x+\dfrac{1}{5}=\dfrac{21}{35}\)
\(x=\dfrac{21}{35}-\dfrac{1}{5}\)
\(x=\dfrac{21}{35}-\dfrac{7}{35}\)
\(x=\dfrac{14}{35}=\dfrac{2}{5}\)

\(x\) + \(\dfrac{1}{5}\) - \(\dfrac{3}{7}\) = \(\dfrac{6}{35}\) 

\(x\) + \(\dfrac{1}{5}\) = \(\dfrac{6}{35}\) + \(\dfrac{3}{7}\)

\(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)

\(x\)       = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)

\(x\) =\(\dfrac{2}{5}\)

Con cảm ơn cô Hoài, cảm ơn bạn Komuro nhé 

Vì \(\left|x-3,5\right|\ge0\); \(\left|4,5-x\right|\ge0\)

=> \(\left|x-3,5\right|+\left|4,5-x\right|\ge0\)

Mà theo đề bài: \(\left|x-3,5\right|+\left|4,5-x\right|=0\)

=> \(\begin{cases}\left|x-3,5\right|=0\\\left|4,5-x\right|=0\end{cases}\)=> \(\begin{cases}x-3,5=0\\4,5-x=0\end{cases}\)=> \(\begin{cases}x=3,5\\x=4,5\end{cases}\), vô lý vì x không thể cùng đồng thời nhận 2 giá trị khác nhau

Vậy không tồn tại giá trị của x thỏa mãn đề bài

làm rồi mà?

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)

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

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

\(\Leftrightarrow2x.2=6x^2+6x+6\)

\(\Leftrightarrow4x=6x^2+6x+6\)

\(\Leftrightarrow6x^2+2x+6=0\)

Ta có \(\Delta=2^2-4.6.6< 0\)

Vậy pt vô nghiệm

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

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

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

\(\Leftrightarrow2.2x=6x^2+6x+6\)\(\Leftrightarrow4x=6x^2+6x+6\)

\(\Leftrightarrow6x^2+2x+6=0\)\(\Leftrightarrow3x^2+x+3=0\)( vô nghiệm vì \(1^2< 4.3.3\)hay \(1< 36\)) 

Vậy tập nghiệm của phương trình là \(S=\varnothing\)