Question

Giải phương trình

a, (x+2)² - x(x+2)=0

b, \(\frac{2x+7}{3}-\frac{x-2}{4}=2\)

c, |x+5|= 3x +1

Giải hộ em với ạ!!! Em đang cần gấp

Answer 1

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

Answer 2

a, -2

b, -2

c, 2

Answer 3

(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)