Question

Tính giá trị của

\(Q=x^2-y^2-z^2-2yz-20x\)biết x+y+z=10.

Giúp với nha

Answer 1

\(Q=x^2-y^2-z^2-2yz-20x\)

=> \(Q=x^2-y^2-z^2-2yz-20x+100-100\)

=> \(Q=\left(x^2-20x+100\right)-\left(y^2+2yz+z^2\right)-100\)

=> \(Q=\left(x-10\right)^2-\left(y+z\right)^2-10^2\)

=> \(Q=\left(x-10-y-z\right)\left(x-10+y+z\right)-10^2\)

=> \(Q=\left[x-\left(x+y+z\right)-y-z\right]\left[x-\left(x+y+z\right)+y+z\right]-10^2\)

=> \(Q=\left[x-\left(x+y+z\right)-y-z\right]\left[x-\left(x+y+z\right)+y+z\right]-10^2\)

=> \(Q=(-2y-2z).0-10^2\)

=> \(Q=0-10^2\)

=> \(Q=-100\)

Answer 2

TA có :

\(x+y+z=10\Rightarrow\left(x+y+z\right)^2=100\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\)\(Q=x^2-y^2-z^2-2yz-20x\)

\(\Rightarrow100+Q=x^2+y^2+z^2+2xy+2yz+2xz+x^2-y^2-z^2-2yz-20x\)\(=2x^2+2xy+2xz-20x\)

\(=2x\left(x+y+z-10\right)\)

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

\(\Rightarrow100+Q=0\Rightarrow Q=-100\)