Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Phân tích thành nhân tử
\(=\left(my+nx\right)\left(ny+mx\right)\)
mn(x2 +y2) +xy(m2 +n2)= mnx2 +mny2 +xym2 +xyn2
=mx(nx + my) +ny( my +nx)
=(mx+ny)(nx+my)
\(x^2-xy-8x+8y\)
\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)
\(mn\left(x^2+y^2\right)+xy\left(m^2+n^2\right)\)
\(=mnx^2+mny^2+xym^2+xyn^2\)
\(=\left(mnx^2+xyn^2\right)+\left(mny^2+xym^2\right)\)
\(=xn\left(mx+ny\right)+ym\left(ny+xm\right)\)
\(=\left(xn+ym\right)\left(mx+ny\right)\)
\(x^2+xy-2y^2=x^2-xy+2xy-2y^2\)
\(=x\left(x-y\right)+2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x+2y\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a: \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=x^2y^2+a^2b^2+x^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(x^2+a^2\right)\)