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B, => 2x+5=3x-8
2x-3x=-8-5
-x=-13
=>x=13
hoặc 2x+5=-3x+8
2x+3x=8-5
5x=3
x=\(\frac{3}{5}\)
\(\left|x+1\right|+\left|x-5\right|=7.\)
\(Th1:x+1< 0;x-5< 0\)
\(x< -1;x< 5\Rightarrow x< -1\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(-\left(x+1\right)-\left(x-5\right)=7\)
\(-x-1-x+5=7\)
\(-2x+4=7\)
\(-2x=3\)
\(x=-1,5\left(tm\right)\)
\(Th2:x+1>0;x-5>0\)
\(x>-1;x>5\Rightarrow x>5\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(x+1+x-5=7\)
\(2x-4=7\)
\(2x=11\)
\(x=5,5\)\(\left(tm\right)\)
\(Th3:x+1\le0;x-5>0\)
\(x\le-1;x>5\)(không xảy ra)
\(Th4:x+1>0;x-5\le0\)
\(x>-1;x\le5\Rightarrow-1< x\le5\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(-\left(x+1\right)+x-5=7\)
\(-x-1+x-5=7\)( không xảy ra)
Vậy x = -1,5 hoặc x = 5,5
\(\)
a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2