Toán phân tích đa thức thành nhân tử (x-y)^3+(y+z)^3-(x-z)^3 17/07/2021 By Valentina phân tích đa thức thành nhân tử (x-y)^3+(y+z)^3-(x-z)^3
$↓$ $(x-y)³+(y+z)³-(x-z)³$ $⇔ x³-3x²y+3xy²-y³+y³+3y²z+3yz²+z³-(x³-3x²z+3xz²-z³)$ $⇔ x³-3x²y+3xy²+3y²z+3yz²+z³+3x²z-3xz²+z³$ $⇒ -3x²y+3xy²+3y²z+3yz²+2z³+3y²z+3x²z-3xz²$Hc tốt!!!! Trả lời
(x−y)³+(y+z)³−(x−z)³(x−y)³+(y+z)³−(x−z)³ ⇔x³−3x²y+3xy²−y³+y³+3y²z+3yz²+z³−(x³−3x²z+3xz²−z³) ⇔x³−3x²y+3xy²−y³+y³+3y²z+3yz²+z³−(x³−3x²z+3xz²−z³) ⇔x³−3x²y+3xy²+3y²z+3yz²+z³+3x²z−3xz²+z³ ⇔x³−3x²y+3xy²+3y²z+3yz²+z³+3x²z−3xz²+z³ ⇒−3x²y+3xy²+3y²z+3yz²+2z³+3y²z+3x²z−3xz² ⇒−3x²y+3xy²+3y²z+3yz²+2z³+3y²z+3x²z−3xz² Xin câu trả lời hay nhất ạ Trả lời
$↓$
$(x-y)³+(y+z)³-(x-z)³$
$⇔ x³-3x²y+3xy²-y³+y³+3y²z+3yz²+z³-(x³-3x²z+3xz²-z³)$
$⇔ x³-3x²y+3xy²+3y²z+3yz²+z³+3x²z-3xz²+z³$
$⇒ -3x²y+3xy²+3y²z+3yz²+2z³+3y²z+3x²z-3xz²$
Hc tốt!!!!
(x−y)³+(y+z)³−(x−z)³(x−y)³+(y+z)³−(x−z)³
⇔x³−3x²y+3xy²−y³+y³+3y²z+3yz²+z³−(x³−3x²z+3xz²−z³)
⇔x³−3x²y+3xy²−y³+y³+3y²z+3yz²+z³−(x³−3x²z+3xz²−z³)
⇔x³−3x²y+3xy²+3y²z+3yz²+z³+3x²z−3xz²+z³
⇔x³−3x²y+3xy²+3y²z+3yz²+z³+3x²z−3xz²+z³
⇒−3x²y+3xy²+3y²z+3yz²+2z³+3y²z+3x²z−3xz²
⇒−3x²y+3xy²+3y²z+3yz²+2z³+3y²z+3x²z−3xz²
Xin câu trả lời hay nhất ạ