tìm x
a, \(\left(x+2\right)^2-x\left(x+3\right)=2\)
b, \(\left(x+2\right)\left(x-2\right)-\left(x+1\right)^2=7\)
c, \(6x^2-\left(2x+1\right)\left(3x-2\right)=1\)
d, \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+1\right)=2\)
e, 6(x-1)(x+1) - (2x-1)(3x+2) + 3 = 0
f, \(x\left(3x+1\right)+\left(x-1\right)^2-\left(2x+1\right)\left(2x-1\right)=0\)
g, \(\left(x+1\right)^3+\left(2-x\right)^3-9\left(x-3\right)\left(x+3\right)=0\)
h, \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3x^2=25\)
a: \(\left(x+2\right)^2-x\left(x+3\right)=2\)
=>\(x^2+4x+4-x^2-3x=2\)
=>x+4=2
=>x=2-4=-2
b: \(\left(x+2\right)\left(x-2\right)-\left(x+1\right)^2=7\)
=>\(x^2-4-\left(x^2+2x+1\right)=7\)
=>\(x^2-4-x^2-2x-1=7\)
=>-2x-5=7
=>-2x=5+7=12
=>x=-6
c: \(6x^2-\left(2x+1\right)\left(3x-2\right)=1\)
=>\(6x^2-\left(6x^2-4x+3x-2\right)=1\)
=>\(6x^2-\left(6x^2-x-2\right)=1\)
=>\(6x^2-6x^2+x+2=1\)
=>x+2=1
=>x=1-2=-1
d: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+1\right)=2\)
=>\(x^2+5x+6-\left(x^2-x-2\right)=2\)
=>\(x^2+5x+6-x^2+x+2=2\)
=>6x+8=2
=>6x=2-8=-6
=>x=-1
e: \(6\left(x-1\right)\left(x+1\right)-\left(2x-1\right)\left(3x+2\right)+3=0\)
=>\(6\left(x^2-1\right)-\left(6x^2+4x-3x-2\right)+3=0\)
=>\(6x^2-6-6x^2-x+2+3=0\)
=>-x-1=0
=>x+1=0
=>x=-1
f: \(x\left(3x+1\right)+\left(x-1\right)^2-\left(2x+1\right)\left(2x-1\right)=0\)
=>\(3x^2+x+x^2-2x+1-\left(4x^2-1\right)=0\)
=>\(4x^2-x+1-4x^2+1=0\)
=>-x+2=0
=>-x=-2
=>x=2
g: \(\left(x+1\right)^3+\left(2-x\right)^3-9\left(x-3\right)\left(x+3\right)=0\)
=>\(x^3+3x^2+3x+1+8-6x+12x^2-x^3-9\left(x^2-9\right)=0\)
=>\(15x^2-3x+9-9x^2+81=0\)
=>\(6x^2-3x+90=0\)
=>\(2x^2-x+30=0\)
\(\Delta=\left(-1\right)^2-4\cdot2\cdot30=1-8\cdot30=1-240=-239< 0\)
=>Phương trình vô nghiệm
h: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3x^2=25\)
=>\(x^3-3x^2+3x-1-\left(x^3+27\right)+3x^2=25\)
=>\(x^3+3x-1-x^3-27=25\)
=>3x-28=25
=>3x=25+28=53
=>\(x=\dfrac{53}{3}\)
