phân tích đa thức thành nhân tử: a> 10x+15y; b> x^2-2xy-4+y^2; c> x(x+y)-3x-3y
giúp mình nhaaaaaaa:>
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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
= (x - y).(x + y) - (15x - 15y)
= (x - y).(x + y) - 15.(x - y)
(x - y).(x + y -15)
\(a,x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\\ b,2x^3-8x=2x\left(x^2-4\right)=2x\left(x-2\right)\left(x+2\right)\\ c,2x^2+6x=2x\left(x+3\right)\\ d,10x+15y=5\left(2x+3y\right)\)
Tick plz
a. \(x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
b. \(2x^3-8x=2x\left(x^2-4\right)=2x\left(x-2\right)\left(x+2\right)\)
c. \(2x^2+6x=2x\left(x+3\right)\)
d. \(10x+15y=5\left(2x+3y\right)\)
a,\(x^3-3x^2+3x-1-y^3=\left(x^3-1\right)-\left(3x^2-3x\right)-y^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)-y^3\)
\(=\left(x-1\right)\left(x^2-2x+1\right)-y^3\)
\(=\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
....
a) 10x + 15y = 5(2x + 3y)
b) x2 - 2xy - 4 + y2
= (x2 - 2xy + y2) - 4
= (x - y)2 - 22
= (x - y + 2)(x - y - 2)
c) x(x + y) - 3x - 3y
= x(x + y) -3(x + y)
= (x - 3)(x + y)
a, \(10x+15y=5\left(2x+3y\right)\)
b, \(x^2-2xy-4+y^2=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
c, \(x\left(x+y\right)-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\)