Question

[tran van binh]

Cho x²-y²-z²=0

C/m (5x-3y+4z).(5x-3y-4z)=(3x-5y)²

Answer 1

[Thắng Nguyễn]

Cách 1:x²-y²-z²=0

=>x²=y²+z²

(5x-3y+4z)(5x-3y-4z)

=(5x-3y)²-16z²

=25x²-30xy+9y²-16z²(*)

Vì x²=y²+z²=>z²=x²-y² nên (*)=25x²-30xy+9y²-16(x²-y²)=(3x-5y)²

Cách 2: cách này dễ hiểu hơn

x²-y²-z²=0

=>x²=y²+z²

(5x-3y+4z).(5x-3y-4z)=(3x-5y)²

<=>(5x-3y)²-16z²=(3x-5y)²

<=>(5x-3y)²-(3x-5y)²=16z²

<=>(8x-8y)(2x+2y)=16z²

<=>16(x²-y²)=16z²

<=>x²=y²+z² (đúng với gt)

Answer 2

[Nguyễn Quỳnh Chi]

Ta có: (5x-3y+4z)(5x-3y-4z)=(5x-3y)^2-16z^2=25x^2-30xy+9y^2-16(x^2-y^2)=25x^2-30xy+9y^2-16x^2+16y^2

                                                                                                            =9x^2-30xy+25y^2=(3x-5y)^2  (đpcm)