Phân tích đa thức thành nhân tử 2 x y - x 2 - y 2 + 16
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(\left(x+y\right)^2-16\)
\(=\left(x+y\right)^2-4^2\)
\(=\left[\left(x+y\right)-4\right]\left[\left(x+y\right)+4\right]\)
\(=\left(x+y-4\right)\left(x+y+4\right)\)
\(=x^2-\left(y-4\right)^2\)
\(=\left(x-y+4\right)\left(x+y-4\right)\)
\(=x^2-\left(y^2-8y+16\right)=x^2-\left(y-4\right)^2=\left(x-y+4\right)\left(x+y-4\right)\)
\(x^2-y^2+10x-6y+16\)
\(=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)
\(=\left(x+5\right)^2-\left(y+3\right)^2\)
\(=\left(x+5-y-3\right)\left(x+5+y+3\right)\)
\(=\left(x-y+2\right)\left(x+y+8\right)\)
\(=\left[4x-4y\right]^2-\left(5x+5y\right)^2\)
\(=\left(4x-4y+5x-5y\right)\left(4x-4y-5x+5y\right)\)
\(=\left(9x-9y\right)\left(y-x\right)\)
\(=-9\left(x-y\right)^2\)
16(x-y)2-25(x+y)2
=16(x-y)2-25(x-y)2
=(16-25)(x-y)2
=-9(x-y)2
(4(x-y))2 - (5(x+y))2 = (4(x-y) +5(x+y))(4(x-y) -5(x+y))
(4x-4y +5x+5y)(4x-4y-5x-5y)
= -(9x +y)(x+9y)