\(a.\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\\\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\\\Leftrightarrow x^2-x^2+4x-3x-6x+8x=+12-4+16=0\\\Leftrightarrow 3x=24\\\Leftrightarrow x=8\)

Vậy \(x=8\) để \(A=B\)

\(b.\left(2+x\right)\left(x-2\right)+3x^2=\left(2x+1\right)^2+2x\\\Leftrightarrow x^2-4+3x^2=4x^2+4x+1+2x\\ \Leftrightarrow x^2+3x^2-4x^2-4x-2x=4+1\\ \Leftrightarrow-6x=5\\\Leftrightarrow x=-\frac{5}{6}\)

Vậy \(x=-\frac{5}{6}\) để \(A=B\)

Bạn xem lại đề b nhé

\(c.\left(x-1\right)\left(x^2+x+1\right)-2x=x\left(x-1\right)\left(x+1\right)\\\Leftrightarrow x^3-1-2x=x\left(x^2-1\right)\\ \Leftrightarrow x^3-1-2x=x^3-x\\\Leftrightarrow x^3-x^3-2x+x=1\\\Leftrightarrow -x=1\\\Leftrightarrow x=-1\)

Vậy \(x=-1\) để \(A=B\)

a) A= (x-3)(x+4)-2(3x-2) và B= (x-4)²

ta có:

(x−3)(x+4)−2(3x−2)= (x-4)²

=> \(x^2-5x-8=x^2-8x+16\)

=>3x=24

=>x=8

b) A= (x+2)(x-2)+3x² và B= (2x+1)² +2x

Ta có:

\(\left(x+2\right)\left(x-2\right)+3x^2=\left(2x+1\right)^2+2x\)

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

=>6x=−5

=>\(x=-\frac{5}{6}\)

c) A= (x-1)(x²+x+1)-2x và B= x(x-1)(x+1)

ta có:

\(\left(x-1\right)\left(x^2+x+1\right)-2x=x\left(x+1\right)\left(x+1\right)\)

=>\(x^3-1-2x=x\left(x^2-1\right)\)

=>\(x^3-1-2x=x^3-x\)

=>\(x=-1\)

d) A= (x+1)³ -(x-2)³ và B= (3x-1)(3x+1)

ta có:

\(\left(x+1\right)^3-\left(x-2\right)^3=\left(3x-1\right)\left(3x+1\right)\)

=>\(9x^2+9x+9=9x^2-1\)

=>\(9x=10\)

=>x=\(\frac{10}{9}\)

chúc bạn học tốt

#Nghi

a: A(x)+B(x)

=5x^3-2x+3x^2+2x-1

=5x^3+3x^2-1

b: A(x)-C(x)

=5x^3-2x-2x^3+3x^2-3x-1

=3x^3+3x^2-5x-1

c: M(x)=B(x)+C(x)

=3x^2+2x-1+2x^3-3x^2+3x+1

=2x^3+5x

d: B(1/3)=3*1/9+2*1/3-1=1/3+2/3-1=0

=>x=1/3 là nghiệm của B(x)