(5x + 1)^2 − (2xy − 3)^2 phân tích da thức thành nhân tử
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=(x^5-5x^2)-(2xy+10y)
=(x^5-5x^2)-(2xy-10y)
=x^2.(x-5)-2y.(x-5)
=(x^2-2y).(x-5)
k nha
\(x^3-5x^2-2xy+10y=x^2\left(x-5\right)-2y\left(x-5\right)=\left(x-5\right)\left(x^2-2y\right)=\left(x-5\right)\left(x-2y\right)\left(x+2y\right)\)
\(=2xy+5x+4y^2+10y\)
\(=x\left(2y+5\right)+2y\left(2y+5\right)\)
\(=\left(x+2y\right)\left(2y+5\right)\)
\(=2x\left(x+2y\right)+5\left(x+2y\right)=\left(x+2y\right)\left(2x+5\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(x^2-5x+2xy-10y\)
\(=\left(x^2-5x\right)+\left(2xy-10y\right)\)
\(=x\left(x-5\right)+2y\left(x-5\right)\)
\(=\left(x+2y\right)\left(x-5\right)\)
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10=\left(x+y\right)^2+7\left(x+y\right)+10\)
=\(\left(x+y+2\right)\left(x+y+5\right)\)
^^
4(x2+xy+xy+y2)+5x+=5y+1
=4(x2+xy+xy+y2)+4x+4y+x+y+1
=4(x2+xy+xy+y2+x+y)+(x+y+1)
=4(x+y)(x+y+1)+(x+y+1)
=(x+y+1)(4x+4y+1)
Hok tốt
\(4\left(x^2+2xy+y^2\right)+5x+5y+1\)
\(\Rightarrow4\left(x+y\right)^2+5\left(x+y\right)+1\)
\(\Rightarrow\left(x+y\right)\left[4\left(x+y\right)+5+1\right]\)
\(\Rightarrow\left(x+y\right)\left(4\left(x+y\right)+6\right)\)
\(\Rightarrow2\left(x+y\right)\left(2\left(x+y\right)+3\right)\)
\(\left(5x+1\right)^2-\left(2xy-3\right)^2\\ =\left[\left(5x+1\right)-\left(2xy-3\right)\right]\left[\left(5x+1\right)+\left(2xy-3\right)\right]\\ =\left(5x+1-2xy+3\right)\left(5x+1+2xy-3\right)\\ =\left(5x-2xy+4\right)\left(5x+2xy-2\right)\)
(5\(x\) + 1)2 - (2\(xy\) - 3)2
= [(5\(x\) + 1) - (2\(xy\) - 3)].[(5\(x\) + 1) + (2\(xy\) - 3)]
= [ 5\(x\) + 1 - 2\(xy\) + 3][5\(x\) + 1 + 2\(xy\) - 3]
= [5\(x\) - 2\(xy\) + (1 + 3)][5\(x\) + 2\(xy\) - (3 - 1)]
= [5\(x\) - 2\(x\)\(y\) + 4][5\(x+2xy\) - 2]