Câu hỏi của Lizy

Question

Phân tích thành nhân tử (mọi người làm chi tiết ạ)$2ax-bx+3cx-2a+b-3c$$ax-bx-2cx-2a+2b+4c$$3ax^2 +3bx^2 +ax+bx+5a+5b$$ax^2 -bx^2 -2ax+2bx-3a+3b$

Shinichi Kudo · Accepted Answer

Câu trả lời:

Hà Quang Minh · Answer

$2ax-bx+3cx-2a+b-3c\
=x\left(2a-b+3c\right)-\left(2a-b+3c\right)\
=\left(x-1\right)\left(2a-b+3c\right)$ $ax-bx-2cx-2a+2b+4c\
=x\left(a-b-2c\right)-2\left(a-b-2c\right)\
=\left(x-2\right)\left(a-b-2c\right)$ $3ax^2+3bx^2+ax+bx+5a+5b\
=3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\
=\left(3x^2+x+5\right)\left(a+b\right)$ $ax^2-bx^2-2ax+2bx-3a+3b\
=x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\
=\left(x^2-2x-3\right)\left(a+b\right)\
=\left(x+1\right)\left(x-3\right)\left(a+b\right)$Câu trả lời: $2ax-bx+3cx-2a+b-3c\
=x\left(2a-b+3c\right)-\left(2a-b+3c\right)\
=\left(x-1\right)\left(2a-b+3c\right)$ $ax-bx-2cx-2a+2b+4c\
=x\left(a-b-2c\right)-2\left(a-b-2c\right)\
=\left(x-2\right)\left(a-b-2c\right)$ $3ax^2+3bx^2+ax+bx+5a+5b\
=3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\
=\left(3x^2+x+5\right)\left(a+b\right)$ $ax^2-bx^2-2ax+2bx-3a+3b\
=x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\
=\left(x^2-2x-3\right)\left(a+b\right)\
=\left(x+1\right)\left(x-3\right)\left(a+b\right)$

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