<=> (3x+2) [(3x+2)+(4-x)] = 0
<=> (3x+2) (3x+2+4-x) = 0\
<=> (3x+2) (2x+6) = 0
<=> 3x+2 = 0 hoặc 2x+6 = 0
3x+2 = 0 <=> 3x=-2 <=> x=-2/32x+6=0 <=> 2x=-6 <=> x=-3Vậy S={-2/3; -3}
<=> (3x+2) [(3x+2)+(4-x)] = 0
<=> (3x+2) (3x+2+4-x) = 0\
<=> (3x+2) (2x+6) = 0
<=> 3x+2 = 0 hoặc 2x+6 = 0
3x+2 = 0 <=> 3x=-2 <=> x=-2/32x+6=0 <=> 2x=-6 <=> x=-3Vậy S={-2/3; -3}
Bằng cách phân tích vế trái thành nhân tử, giải các PT sau:
a) \(2x.\left(x-3\right)+5\left(x-3\right)\)
b) \(\left(x^2-4\right)+\left(x-2\right).\left(3-2x\right)=0\)
c) \(x^3-3x^2+3x-1=0\)
Giải PT sau
\(\left(x^2+3x-4\right)^3+\left(2x^2-5x+3\right)^3=\left(3x^2-2x-1\right)^3\)
Giải pt sau :
\(\left(3x-2\right)\left[\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right]=0\)
Giải các pt sau
a, \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)\)=0
b,\(\left(2x-1\right)\left(x-5\right)\left(x^2+3\right)\)=0
c,\(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
d,\(\left(2x+3\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
giải pt
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
b)\(\left(x-4\right)^2-\left(x+2\right)\left(x-6\right)=0\)
c)\(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)
Giải các phương trình sau:
1, \(\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\)
2, \(\left(x-2\right)\left(2x-1\right)=x^2-2x\)
3, \(3x^2-4x+1=0\)
4, \(\left|2x-4\right|=0\)
5, \(\left|3x+2\right|=4\)
6, \(\left|2x-5\right|=\left|-x+2\right|\)
*Giúp mình với mình đg cần gấp ạ T_T
giải pt:
a) \(\left(5x+1\right)^2=\left(3x-2\right)^2\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
c) \(\left(x+3\right)^4+\left(x+5\right)^4=2\)
d) \(x^4-3x^3+4x^2-3x+1=0\)
Giải PT: \(x\left(x+2\right)\left(x^3+3x^2+3x+1\right)+1=0\)
giải pt sau
a)\(\frac{3}{7}x-1=\frac{1}{7}x\left(3x-7\right)\)
b)\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)