a)
\(A\left(x\right)=3\left(x^2+2-4x\right)-2x\left(x-2\right)+17\\ =3x^2+6-12x-2x^2+4x+17\\ =\left(3x^2-2x^2\right)+\left(-12x+4x\right)+\left(6+17\right)\\ =x^2-8x+23\)
\(B\left(x\right)=3x^2-7x+3-3\left(x^2-2x+4\right)\\ =3x^2-7x+3-3x^2+6x-12\\ =\left(3x^2-3x^2\right)+\left(-7x+6x\right)+\left(3-12\right)\\ =-x-9\)
b)
\(N\left(x\right)-B\left(x\right)=A\left(x\right)\\ \Rightarrow N\left(x\right)=A\left(x\right)+B\left(x\right)\\ =\left(x^2-8x+23\right)+\left(-x-9\right)\\ =x^2-8x+23-x-9\\ =x^2-9x+14\)
\(A\left(x\right)-M\left(x\right)=B\left(x\right)\\ \Rightarrow M\left(x\right)=A\left(x\right)-B\left(x\right)\\ =\left(x^2-8x+23\right)-\left(-x-9\right)\\ =x^2-8x+23+x+9\\ =x^2-7x+32\)
c) Thay `x=2` vào N(x) ta có:
\(N\left(2\right)=2^2-9\cdot2+14=0\)
=> x=2 là nghiệm của N(x)
\(N\left(x\right)=x^2-9x+14\\ =x^2-2x-7x+14\\ =x\left(x-2\right)-7\left(x-2\right)\\ =\left(x-7\right)\left(x-2\right)\)
=> Nghiệm còn lại là: x - 7 = 0 => x = 7
d) Thay `x=2/3` vào A(x) ta có:
\(A\left(x\right)=\left(\dfrac{2}{3}\right)^2-8\cdot\left(\dfrac{2}{3}\right)+23=\dfrac{4}{9}-\dfrac{16}{3}+23=\dfrac{4}{9}-\dfrac{48}{9}+\dfrac{207}{9}=\dfrac{163}{9}\)
