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a) 15x + 15y = 15(x + y)
b) 6x - 10y = 2(3x - 5y)
c) 2a + 4b - 6c = 2(a + 2b - 3c)
d) 6xy - 12x - 18y = 6(xy - 2x - 3y)
e) 2(x + y) - 5a(x + y) = (2 - 5a)(x + y)
f) 6x(x - y) + 5(y - x) = 6x(x - y) + (-5)(x - y) = (6x - 5)(x - y)
a) \(15x+15y=15\left(x+y\right)\)
b) \(6x-10y=2\left(3x-5y\right)\)
c) \(2a+4b-6c=2\left(a+2b-3c\right)\)
d) \(6xy-12x-18y=6\left(xy-2x-3y\right)\)
e) \(2\left(x+y\right)-5a\left(x+y\right)=\left(2-5a\right)\left(x+y\right)\)
f) \(6x\left(x-y\right)+5\left(y-x\right)=6x\left(x-y\right)-5\left(x-y\right)=\left(6x-5\right)\left(x-y\right)\)
Đa thức này không phân tích được thành nhân tử
Nó phân tích được khi đề là: \(x^2-6y-y^2-9\) hoặc \(x^2-6x-y^2+9\)
\(=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
a, x^2 - x - y^2 - y
= (x^2 - y^2) - ( x+ y)
= ( x- y)(x+y) - ( x+y )
= ( x - y - 1 )(x+ y)
b, x^2+ 6x + 9 - y^2
= ( x+ 3)^2 - y^2
= ( x+ 3 -y)( x + 3 +y)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
a: =(x-y)(5-y)
b: \(=x^2-6x+9-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
\(a,5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\\ b,x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)