a.(x+2)2-x(x+2)=0
\(\Leftrightarrow\)(x+2)(x-2-x)=0
\(\Leftrightarrow\)(x+2)*2=0
\(\Leftrightarrow\)x+2=0
\(\Leftrightarrow\)x=-2
vay s={-2}
b.\(\frac{2x+7}{3}\)-\(\frac{x-2}{4}\)=2
\(\Leftrightarrow\)\(\frac{4\left(2x+7\right)}{12}\)+\(\frac{-3\left(x-2\right)}{12}\)=\(\frac{24}{12}\)
\(\Leftrightarrow\)8x+28-3x+6=24
\(\Leftrightarrow\)5x=-10
\(\Leftrightarrow\)x=-2
vay s={-2}
c.|x+5|=3x+1
neu x+5\(\ge\)0 thi |x+5|=x+5
\(\Leftrightarrow\)x\(\ge\)-5
ta co phuong trinh
x+5=3x+1
\(\Leftrightarrow\)-2x=-4
\(\Leftrightarrow\)x=2( thoa man dieu kien x\(\ge\)-5)
neu x+5<0 thi |x+5|=5-x
\(\Leftrightarrow\)x<-5
ta co phuong trinh
5-x=3x+1
\(\Leftrightarrow\)-4x=-4
\(\Leftrightarrow\)x=1 (k thoa man dieu kien x<5)
vay s={2}
chuc bn hoc tot
(x+2)(x+2-x)=0
(x+2). 2 = 0
<=> x +2 = 0
<=> x = -2
b , \(\frac{2x+7}{3}-\frac{x-2}{4}=2\)
\(\frac{4\left(2x+7\right)}{12}-\frac{3\left(x-2\right)}{12}=\frac{24}{12}\)
=> 8x + 28 - 3x + 6 = 24
<=> 5x = -10
=> x = -2
c , \(|x+5|=3\)x +1
\(|x+5|=x+5\) <=> x + 5 \(\ge0\)
<=> x \(\ge\) -5
\(|x+5|=3\)
\(x+5=3\)
x =-2 (tm)
+ \(|x+5|=-3\Leftrightarrow x+5< 0=>x< -5\)
\(|x+5|=-3\)
x+5 = -3 x=-8 (tm)
