Câu hỏi của Nguyễn Bảo Hân

Question

Rút gọn biểu thức sau4( 3x + y)'2 + (x - y)'25) -(x + 5)'2 - (x - 3)'26) (3x - 2)'2 - (3x - 1)'2

HT.Phong (9A5) · Accepted Answer

$4)\left(3x+y\right)^2+\left(x-y\right)^2\
=9x^2+6xy+y^2+x^2-2xy+y^2\
=10x^2+4xy+2y^2\
5)-\left(x+5\right)^2-\left(x-3\right)^2\
=-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\
=-x^2-10x-25-x^2+6x-9\
=-2x^2-4x-34
\
6)\left(3x-2\right)^2-\left(3x-1\right)^2\
=\left(9x^2-12x+4\right)-\left(9x^2-6x+1\right)\
=9x^2-12x+4-9x^2+6x-1\
=-6x+3$Câu trả lời: $4)\left(3x+y\right)^2+\left(x-y\right)^2\
=9x^2+6xy+y^2+x^2-2xy+y^2\
=10x^2+4xy+2y^2\
5)-\left(x+5\right)^2-\left(x-3\right)^2\
=-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\
=-x^2-10x-25-x^2+6x-9\
=-2x^2-4x-34
\
6)\left(3x-2\right)^2-\left(3x-1\right)^2\
=\left(9x^2-12x+4\right)-\left(9x^2-6x+1\right)\
=9x^2-12x+4-9x^2+6x-1\
=-6x+3$

Nguyễn Lê Phước Thịnh · Answer

4: $\left(3x+y\right)^2+\left(x-y\right)^2$$=9x^2+6xy+y^2+x^2-2xy+y^2$$=10x^2+4xy+2y^2$5: $-\left(x+5\right)^2-\left(x-3\right)^2$$=-x^2-10x-25-x^2+6x-9$$=-2x^2-4x-34$6: $\left(3x-2\right)^2-\left(3x-1\right)^2$$=\left(3x-2-3x+1\right)\left(3x-2+3x-1\right)$$=-1\left(6x-3\right)=-6x+3$Câu trả lời: 4: $\left(3x+y\right)^2+\left(x-y\right)^2$$=9x^2+6xy+y^2+x^2-2xy+y^2$$=10x^2+4xy+2y^2$5: $-\left(x+5\right)^2-\left(x-3\right)^2$$=-x^2-10x-25-x^2+6x-9$$=-2x^2-4x-34$6: $\left(3x-2\right)^2-\left(3x-1\right)^2$$=\left(3x-2-3x+1\right)\left(3x-2+3x-1\right)$$=-1\left(6x-3\right)=-6x+3$

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