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NoreksVenetor
@NoreksVenetor
October 2021
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Помогите пожалуйста, надо разложить на множители выражение
х((у+z) в квадрате) + y((x+z) в квадрате) + z((x+y) в квадрате) - 4xyz
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nafanya2014
Verified answer
х(у+z)²+ y(x+z)² + z(x+y)² - 4xyz
=
=х(у+z)²+ y(x+z)² + z(x²+2xy+y²) - 4xyz=
=х(у+z)²+ y(x+z)² + z(x²-2xy+y²) =
=х(у+z)²+ y(x+z)² + z(x-y)² =
=х(у+z)²+ y(x-y+y+z)² + z(x-y)² =
=х(у+z)²+ y(x-y)²+2y(x-y)(y+z)+y(y+z)² + z(x-y)² =(y+z)²(x+y)+(x-y)²(y+z)+2y(x-y)(y+z)=
=(y+z)·((y+z)(x+y)+(x-y)²+2y(x-y))=(y+z)·((y+z)(x+y)+(x-y)(x-y+2y))=
(y+z)·((y+z)(x+y)+(x-y)(x+y))=(y+z)(x+y)(y+z+x-y)=(y+z)(x+y)(z+x)
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Answers & Comments
Verified answer
х(у+z)²+ y(x+z)² + z(x+y)² - 4xyz==х(у+z)²+ y(x+z)² + z(x²+2xy+y²) - 4xyz=
=х(у+z)²+ y(x+z)² + z(x²-2xy+y²) =
=х(у+z)²+ y(x+z)² + z(x-y)² =
=х(у+z)²+ y(x-y+y+z)² + z(x-y)² =
=х(у+z)²+ y(x-y)²+2y(x-y)(y+z)+y(y+z)² + z(x-y)² =(y+z)²(x+y)+(x-y)²(y+z)+2y(x-y)(y+z)=
=(y+z)·((y+z)(x+y)+(x-y)²+2y(x-y))=(y+z)·((y+z)(x+y)+(x-y)(x-y+2y))=
(y+z)·((y+z)(x+y)+(x-y)(x+y))=(y+z)(x+y)(y+z+x-y)=(y+z)(x+y)(z+x)