x= 3m-3/m-2

Tại m =2 thì pt vô nghiệm 

Tại m khác 2 thì có nghiệm duy nhất vì đây là hàm bậc nhất

Ko cần biet vi ko biet ang ang

\(\dfrac{1}{2022}\) \(\times\) \(\dfrac{2}{5}\) + \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{7}{5}\) - \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{8}{10}\)

= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{2}{5}\) + \(\dfrac{7}{5}\) - \(\dfrac{8}{10}\))

= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{9}{5}\) - \(\dfrac{4}{5}\))

= \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{5}{5}\)

=  \(\dfrac{1}{2022}\times1\)

= \(\dfrac{1}{2022}\)

c la lon nhat 

làm ơn giải rõ giúp mk vì sao c lớn nhất

`|x-2|=2x-3(x>=3/2)`

`<=>` \(\left[ \begin{array}{l}x-2=2x-3\\x-2=3-2x\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=1(l)\\3x=5\end{array} \right.\) 

`<=>x=5/3(Tm(`

`2)A=-x^2+2x+9`

`=-(x^2-2x)+9`

`=-(x^2-2x+1)+1+9`

`=-(x-1)^2+10<=10`

Dấu "=" xảy ra khi `x=1.`

1,

* \(|x-2|=x-2< =>x\ge2\)

\(=>x-2=2x-3< =>x=1\left(ktm\right)\)

*\(\left|x-2\right|=2-x< =>x< 2\)

\(=>2-x=2x-3< =>x=\dfrac{5}{3}\left(tm\right)\)

vậy x=5/3

2, \(A=-x^2+2x+9=-\left(x^2-2x-9\right)=-\left(x^2-2x+1-10\right)\)

\(=-\left[\left(x-1\right)^2-10\right]=-\left(x-1\right)^2+10\le10\)

dấu"=" xảy ra<=>x=1

Bài 1: 

Ta có: \(\left|x-2\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2x-3\left(x\ge2\right)\\2-x=2x-3\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2x=-3+2\\-x-2x=-3-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=-1\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(lọai\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)

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

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

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

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

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

\(x\) =\(\dfrac{2}{5}\)

Con cảm ơn cô Hoài, cảm ơn bạn Komuro nhé 

\(a)\)

\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)

\(\Leftrightarrow x-x^2+1=3x+1\)

\(\Leftrightarrow x^2-2x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b)\)

\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)

\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)

\(\Leftrightarrow x^2+2x+1=x^2+10\)

\(\Leftrightarrow2x-9=0\)

\(\Leftrightarrow2x=9\)

\(\Leftrightarrow x=\frac{2}{9}\)