phân tích thành nhân tử
x2+3cd(2-3cd)-10xy-1 +25y2
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(x^2-10xy+25y^2)-(1-3cd(2-3cd))
=(x^2-2.x.5.y+(5y)2)-(1-6cd+9(cd)2)
=(x-5)^2-((3cd)2-2.3cd.1+12)
=(x-5)^2-(3cd-1)2
=(x-5+3cd-1).(x-5-3cd+1)
=(x+3cd-6).(x-3cd-4)...
Theo mk là z. Có thể phân tích nữa,đúng hay sai mk cx chưa chắc chắn vì mk cũng mới học thui.
\(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2=x^2+6cd-\left(3cd\right)^2-10xy-1+\left(5y\right)^2\\ \)
\(=x^2-10xy+\left(5y\right)^2-\left(1-6cd+\left(3cd\right)^2\right)\)
\(=\left(x-5y\right)^2+6cd-1-\left(3cd\right)^2=\left(x-5y\right)^2-\left(1-3cd\right)^2\)
\(=\left(x-5y-1+3cd\right)\left(x-5y+1-3cd\right)\)
Câu 1:
\(\left(x+y\right)^2+3\left(x+y\right)-10\)
\(=\left(x+y\right)^2+3\left(x+y\right)+2,25-12,25\)
\(=\left(x+y+1,5\right)^2-3,5^2\)
\(=\left(x+y+1,5+3,5\right)\left(x+y+1,5-3,5\right)\)
\(=\left(x+y+5\right)\left(x+y-2\right)\)
Câu 2:
\(2x^2-y^2+xy\)
\(=2x^2-y^2+2xy-xy\)
\(=\left(2x^2+2xy\right)-\left(xy+y^2\right)\)
\(=2x\left(x+y\right)-y\left(x+y\right)\)
\(=\left(2x-y\right)\left(x+y\right)\)
Câu 3:
\(x^{64}+x^{32}+1\)
\(=x^{64}+2x^{32}+1-x^{32}\)
\(=\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\)
\(=\left(x^{32}+1+x^{16}\right)\left(x^{32}+1-x^{16}\right)\)
\(=\left(x^{32}+x^{16}+1\right)\left(x^{32}-x^{16}+1\right)\)
Câu 4:
\(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2\)
\(=x^2+25y^2-10xy+6cd-\left(3cd\right)^2-1\)
\(=\left(x^2+25y^2-10xy\right)-\left(\left(3cd\right)^2+1-6cd\right)\)
\(=\left(x+5y\right)^2-\left(3cd-1\right)^2\)
\(=\left(\left(x+5y\right)+\left(3cd-1\right)\right)\cdot\left(\left(x+5y\right)-\left(3cd-1\right)\right)\)
\(=\left(x+5y+3cd-1\right)\left(x+5y-3cd+1\right)\)
a)(x-z)^2-y^2+2y-1=(x-z)^2-(y^2-2y+1) b)1-2a+2bc+a^2-b^2-c^2=a^2-2a+1-b^2+2bc-c^2
=(x-z)^2-(y-1)^2 =(a-1)^2-(b^2-2bc+c^2)
=(x-z-y+1)(x-z+y-1) =(a-1)^2-(b-c)^2
=(a-1-b+c)(a-1+b-c)
c)x^2+y^2-2xy-x+y=x^2-2xy+y^2-(x-y)
=(x-y)^2-(x-y)
=(x-y)(x-y-1)
\(a,Sửa:x^2+4xy-9+4y^2=\left(x+2y\right)^2-9=\left(x+2y-3\right)\left(x+2y+3\right)\\ b,=\left(x-6y\right)^2-1=\left(x-6y-1\right)\left(x-6y+1\right)\\ c,=36-\left(x-5y\right)^2=\left(6-x+5y\right)\left(6+x-5y\right)\)
1)
b) \(\left(x-z\right)^2-y^2+2y-1\)
\(=\left(x^2-2xz+z^2\right)-\left(y-1\right)^2\)
\(=\left(y-z\right)^2-\left(y-1\right)^2\)
\(=\left[\left(x-z\right)+\left(y-1\right)\right]\cdot\left[\left(x-z\right)-\left(y+1\right)\right]\)
\(=\left(x-z+y-1\right)\cdot\left(x-z-y-1\right)\)