Question

cho x^2 =y^2+4z^2 chứng minh (5x-3y+8z )(5x-3y-8z)+1 luôn duong

Answer 1

\(x^2=y^2+4z^2\Rightarrow x^2-y^2=4z^2\)

\(A=\left(5x-3y+8x\right)\left(5x-3y-8z\right)+1\)

\(=\left(5x-3y\right)^2-64z^2+1\)

\(=\left(5x-3y\right)^2-16\left(x^2-y^2\right)+1\)

\(=25x^2+9y^2-30xy-16x^2+16y^2+1\)

\(=9x^2-30xy+25y^2+1\)

\(=\left(3x-5y\right)^2+1>0\) \(\forall x;y\)

Answer 2

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