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\(\left(x+y\right)^2+7\left(x+y\right)+10=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10=\left(x+y+2\right)\left(x+y+5\right)\)
a ) x ^ 2 + 2xy + 7x + 7y + y ^2 + 10 = ( x + y ) ^2 + 7 ( x + y ) + 10 = ( x + y ) ( x + y + 17 )
+,
= (x-y)^2 - z.(x-y) = (x-y).(x-y-z)
+,
=(x-y).(x+y)-(x-y) = (x-y).(x+y-1)
+,
=x^3.(x-1)-(x^2-1) = x^3.(x-1).(x-1).(x+1) = (x-1).(x^3-x-1)
+,
=a.(x^2-y^2)-7.(x+y) = a.(x+y).(x-y)-7.(x+y) = (ax+ay-7).(x+y)
\(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-\left(xz-yz\right)\)
\(=\left(x-y\right)\left(x-y\right)-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
\(x^2-y^2-x+y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
\(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
\(ax^2-ay^2-7x-7y\)
\(=a\left(x^2-y^2\right)-\left(7x+7y\right)\)
\(=a\left(x-y\right)\left(x+y\right)-7\left(x-y\right)\)
\(=\left(x-y\right)\left[a\left(x+y\right)-7\right]\)
a)
\(10x^2+10xy+5x+5y\)
\(=10x\left(x+y\right)+5\left(x+y\right)\)
\(=5\left(x+y\right)\left(2x+1\right)\)
b)
\(x^3+x^2-x-1\)
\(=x^2\left(x+1\right)-\left(x+1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
c)
\(x+2a\left(x-y\right)-y\)
\(=\left(x-y\right)+2a\left(x-y\right)\)
\(=\left(x-y\right)\left(2a+1\right)\)
d)
\(x^2-y^2+7x-7y\)
\(=\left(x+y\right)\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+1\right)\)
\(6x^2+xy-7x-2y^2+7y-5=-\left(y-2x-1\right)\left(2y+3x-5\right)\)
Trả lời:
\(6x^2+xy-7x-2y^2+7y-5=-\left(y-2x-1\right)\left(2y+3x-5\right)\)
Hok_tốt nha
a) 7(x-y)
b) (x-3y)2
c) xy(x+y)-(x+y)=(xy-1)(x+y)
d) x2-3x+2x-6=x(x-3)+2(x-3)=(x+2)(x-3)
x^2+2xy+y^2+7x+7y=(x+y)^2+7(x+y)=(x+y)(x+y+7)