a: Đặt \(C=\left(2x+y\right)^2+\left(2x-y\right)^2+2\left(4x^2-y^2\right)+3y\)
\(=\left(2x+y\right)^2+2\cdot\left(2x+y\right)\cdot\left(2x-y\right)+\left(2x-y\right)^2+3y\)
\(=\left(2x+y+2x-y\right)^2+3y=\left(4x\right)^2+3y=16x^2+3y\)
Khi x=2;y=-3 thì \(C=16\cdot2^2+3\cdot\left(-3\right)=64-9=55\)
b: Đặt \(B=\left(x+2\right)\left(x^2-2x+4\right)-x\left(x+4\right)\left(x-4\right)\)
\(=x^3+8-x\left(x^2-16\right)\)
\(=x^3+8-x^3+16x=16x+8\)
Khi x=-1/16 thì \(B=16\cdot\dfrac{-1}{16}+8=-1+8=7\)
c:
Đặt A=\(\left(y-1\right)^3-\left(y-1\right)\left(y^2+y+1\right)+3\left(y-1\right)\left(y+1\right)\)
\(=y^3-3y^2+3y-1-\left(y^3-1\right)+3\left(y^2-1\right)\)
\(=-3y^2+3y+3y^2-3=3y-3\)
Khi \(y=-\dfrac{1}{3}\) vào A, ta được:
\(A=3\cdot\dfrac{-1}{3}-3=-1-3=-4\)