a: 5-(x-6)=4(3-2x)
=>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
=>\(x=\dfrac{1}{7}\)
b: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
=>\(x\left(x+2\right)^2-4x^2=\left(x-2\right)\left(x^2+2x+4\right)\)
=>\(x\left(x^2+4x+4\right)-4x^2-\left(x^3-8\right)=0\)
=>\(x^3+4x^2+4x-4x^2-x^3+8=0\)
=>4x+8=0
=>4x=-8
=>x=-2
c: \(7-\left(2x+4\right)=-\left(x+4\right)\)
=>\(7-2x-4=-x-4\)
=>-2x+3=-x-4
=>-x=-7
=>x=7
d: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
=>\(x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)
=>\(3x^2+12x-9=3x^2+3x+1\)
=>12x-9=3x+1
=>9x=10
=>\(x=\dfrac{10}{9}\)
e: \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)
=>\(2x^2-3x+2x-3=2x^2+10x-x-5\)
=>x-3=9x-5
=>x-9x=-5+3
=>-8x=-2
=>\(x=\dfrac{1}{4}\)
f: \(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
=>\(x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=10x-5x^2-11x-22\)
=>\(x^3-3x^2+3x-1-x^3-2x^2-x=-5x^2-x-22\)
=>\(-5x^2+2x-1=-5x^2-x-22\)
=>2x-1=-x-22
=>3x=-21
=>x=-7
g: \(\left(x-1\right)-\left(2x-1\right)=9-x\)
=>x-1-2x+1=9-x
=>9-x=-x
=>9=0(vô lý)
=>\(x\in\varnothing\)
h: \(\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
=>\(x^2+4x-3x-12-6x+4=x^2-8x+16\)
=>\(x^2-5x-8=x^2-8x+16\)
=>-5x-8=-8x+16
=>3x=24
=>x=8
i: \(x\left(x+3\right)^2-3x=\left(x+2\right)^3+1\)
=>\(x\left(x^2+6x+9\right)-3x=x^3+6x^2+12x+9\)
=>\(x^3+6x^2+6x=x^3+6x^2+12x+9\)
=>12x+9=6x
=>6x=-9
=>x=-3/2
j: \(\left(x+1\right)\left(x^2-x+1\right)-2x=x\left(x+1\right)\left(x-1\right)\)
=>\(x^3+1-2x=x\left(x^2-1\right)\)
=>\(x^3-2x+1=x^3-x\)
=>-2x+1=-x
=>-x=-1
=>x=1