Question

a)A=5x²z-10xyz+5y²z tại x=124,y=24,z=2b)B=2x²+2y²-x²z-y²z+z-2 tại x=1,y=1,z=-1

Answer 1

\(A=5x^2z-10xyz+5y^2z=5z\left(x^2-2xy+y^2\right)=5z\left(x-y\right)^2\)

Thay x = 124, y = 24, z = 2 vào A, ta có:

\(5\times2\times\left(124-24\right)^2=10\times100^2=10\times10000=100000\)

Vậy A = 10 000 khi x = 124, y = 24, z = 2.

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

Thay x = 1, y = 1, z = - 1 vào B, ta có:

\(B=\left[2-\left(-1\right)\right]\left(1^2+1^2-1\right)=3\times1=3\)

Vậy B = 3 khi x = 1, y = 1, z = - 1.