Phân tích đa thức thành nhân tử:
5x2y−xy
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a, Thay x = -2 và y = 1 vào BT, ta được:
\(M=5.\left(-2\right)^2.1+3.\left(-2\right).1-21=20-6-21=-7\)
b, Ta có: \(a^2-a-6=a^2-3a+2a-6=a\left(a-3\right)+2\left(a-3\right)=\left(a-3\right)\left(a+2\right)\)
\(a,5x^2y-10xy^2=5xy\left(x-2y\right)\\ b,x^2+2xy+y^2-5x-5y=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\\ c,x^2-6x+8=\left(x^2-2x\right)-\left(4x-8\right)=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\\ d,5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)
Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.
\(5x^2y+5xy^2-a^2x-a^2y\)
\(=5xy\left(x+y\right)-a^2\left(x+y\right)\)
\(=\left(x+y\right)\left(5xy-a^2\right)\)
a) \(5x^2y-20xy+20y=5y\left(x^2-4x+4\right)=5y\left(x-2\right)^2\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(3x^2y-12y=3y\left(x^2-4\right)=3y\left(x-2\right)\left(x+2\right)\)
d) \(7x^3-28x^2+28x=7x\left(x^2-4x+4\right)=7x\left(x-2\right)^2\)
a: \(5x^2y-20xy+20y\)
\(=4y\left(x^2-4x+4\right)\)
\(=4x\left(x-2\right)^2\)
b: \(3x^3+6x^2+3x\)
\(=3x\left(x^2+2x+1\right)\)
\(=3x\left(x+1\right)^2\)
c: \(3x^2y-12y\)
\(=3y\left(x^2-4\right)\)
\(=3y\left(x-2\right)\left(x+2\right)\)
d: \(7x^3-28x^2+28x\)
\(=7x\left(x^2-4x+4\right)\)
\(=7x\left(x-2\right)^2\)
\(a,=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ b,=\left(x+y\right)\left(x-5\right)\\ c,=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-2y\right)\left(x-y\right)\\ d,=x^2-2xy=x\left(x-2y\right)\\ e,=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(5x^2y-xy=xy\left(5x-1\right)\)
\(5 x 2 y − x y = x y ( 5 x − 1 )\)