\(a,\left(4x+12\right)\left(3x-2\right)-\left(3x-3\right)\left(4x-1\right)=27\\ 12x^2+36x-8x-24-\left(12x^2-12x-3x+3\right)=27\\ 12x^2+28x-24-12x^2-15x-3-27=0\\ 13x-54=0\\ 13x=54\\ x=18\)
\(b,60x^2+35x-60x^2+15x+100=0\\ 50x=-100\\ x=-2\)
a) \(4\left(x+3\right)\left(3x-2\right)-3\left(x-1\right)\left(4x-1\right)=-27\)
\(12x^2-8x+36x-24-12x^2+3x+12x-3=-27\)
\(43x=-27+27\)
\(43x=0\)
\(x=0\)
Vậy \(x=0\)
b) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(60x^2+35x-60x^2+15x=-100\)
\(50x=-100\)
\(x=-\dfrac{100}{50}\)
\(x=-2\)
Vậy \(x=-2\)
c) \(0,6x\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)
\(0,6x^2-0,3x-0,6x^2-0,39x=0,138\)
\(-0,69x=0,138\)
\(x=\dfrac{-0,138}{0,69}\)
\(x=-0,2\)
Vậy \(x=-0,2\)
d) \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)
\(\left(x^2+2x+x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2=27\)
\(17x=27-10\)
\(17x=17\)
\(x=\dfrac{17}{17}\)
\(x=1\)
Vậy \(x=1\)
