TH1 : Xét \(x< -2\)

\(\Leftrightarrow\left|x+2\right|+\left|7-x\right|=3x+4\)

\(\Leftrightarrow-x-2+7-x=3x+4\)

\(\Leftrightarrow-2x+5=3x+4\)

\(\Leftrightarrow x=\frac{1}{5}\)( loại )

TH2 : Xét \(-2< x< 7\)

\(\Leftrightarrow\left|x+2\right|+\left|7-x\right|=3x+4\)

\(\Leftrightarrow x+2+7-x=3x+4\)

\(\Leftrightarrow x=\frac{5}{3}\left(TM\right)\)

TH3 : Xét \(x\ge7\)

\(\Rightarrow x+2+7+x=3x+4\)

\(\Leftrightarrow x=-9\)( loại )

Nếu \(x< -2\)

\(\rightarrow-\left(x+2\right)+\left(7-x\right)=3x+4\)

\(\Leftrightarrow-x-2+7-x-3x-4=0\)

\(\Leftrightarrow-5x=-1\)

\(\Leftrightarrow x=\frac{1}{5}\left(ktm\right)\)

Nếu \(-2\le x\le7\\ \rightarrow\left(x+2\right)+\left(7-x\right)=3x+4\)

\(\Leftrightarrow x+2+7-x=3x+4\)

\(\Leftrightarrow9-4=3x\\ \Leftrightarrow5=3x\)

\(\Leftrightarrow x=\frac{5}{3}\left(tm\right)\)

Nếu \(x>7\)

\(\rightarrow\left(x+2\right)-\left(7-x\right)=3x+4\)

\(\Leftrightarrow x+2-7+x=3x+4\)

\(\Leftrightarrow2x-5=3x+4\\ \Leftrightarrow x=-9\left(ktm\right)\)

Vậy, \(S=\left\{\frac{5}{3}\right\}\)

@Cừu

\({x-5 \over x-8}\)

\(\frac{-3x+5}{2}< 1\Leftrightarrow\frac{-3x+5}{2}-1< 0\)

\(\Leftrightarrow\frac{-3x+5-2}{2}< 0\Leftrightarrow\frac{-3x+3}{2}< 0\)

\(\Rightarrow-3x+3< 0\)vì 2 > 0 

\(\Leftrightarrow-3\left(x-1\right)< 0\Leftrightarrow x-1>0\Leftrightarrow x>1\)

Vậy tập nghiệm của bất phương trình là S = { x | x > 1 }

M N P E F

a) Xét \(\Delta MNP\) và \(\Delta EFP\)có:

\(\widehat{NMP}=\widehat{FEP}\left(=90^0\right)\).

\(\widehat{MPN}\)chung.

\(\Rightarrow\Delta MNP~\Delta EFP\left(g.g\right)\)(điều phải chứng minh).

\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)ĐK : \(x\ne8;9;10;11\)

\(\Leftrightarrow\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)

\(\Leftrightarrow\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)

\(\Leftrightarrow x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\right)=0\Leftrightarrow x=0\)

Vậy tập nghiệm của phương trình là S = { 0 }