Question

Answer 1

a: \(\left(x+1\right)\left(2-3x\right)+3x\left(x-1\right)=10\)

=>\(2x-3x^2+2-3x+3x^2-3x=10\)

=>-4x=8

=>x=-2

b: \(\left(4x-1\right)\left(4x+1\right)-16\left(x+1\right)\left(x-5\right)=15\)

=>\(16x^2-1-16\left(x^2-4x-5\right)=15\)

=>\(16x^2-1-16x^2+64x+80=15\)

=>64x+79=15

=>64x=-64

=>x=-1

c: 

ĐKXĐ: x<>0

\(\dfrac{\left(4x^3+3x^2\right)}{-x^2}+\dfrac{15x^4+6x^3}{3x^3}=0\)

=>\(-4x-3+5x+2=0\)

=>x-1=0

=>x=1(nhận)

d: ĐKXĐ: \(x\notin\left\{0;-\dfrac{1}{2}\right\}\)

\(\dfrac{6x^6-4x^4}{x^4}-\dfrac{\left(2x+1\right)^2}{2x+1}-\dfrac{6x^8}{x^6}=0\)

=>\(6x^2-4-2x-1-6x^2=0\)

=>-2x-5=0

=>2x=-5

=>\(x=-\dfrac{5}{2}\left(nhận\right)\)