\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)\)

\(=\left[a\left(x+y\right)+b\left(x+y\right)\right]\left[a\left(x-y\right)-b\left(x-y\right)\right]\)

\(=\left(a+b\right)\left(a-b\right)\left(x+y\right)\left(x-y\right)\)

\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left(a+b-1\right)\left(a+b+1\right)\left(a-b-3\right)\left(a-b+3\right)\)

a)

(ax+by)² - (ay+bx)²

=(ax+by-ay-bx)(ax+by+ay+bx)

=[ a(x-y) -b(x-y)][ a(x+y) + b(x+y)]

=(a-b)(x-y)(a+b)(x+y)

b)(a²+b²-5)² - 4(ab+2)²

=(a²+b²-5-2ab-4)(a²+b²-5+2ab+4)

=[ (a-b)² -9][ (a+b)² -1]

=(a-b-3)(a-b+3)(a+b-1)(a+b+1)

a, \(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax+ay+bx+by\right)\left(ax-ay+bx-by\right)\)

\(=\left(a+b\right)\left(x+y\right)\left(a-b\right)\left(x-y\right)\)

b, \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b+1\right)\left(a+b-1\right)\)

\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)

\(=\left[a^2-\left(b^2-2bc+c^2\right)\right].\left[\left(b^2+2bc+c^2\right)-a^2\right]\)

\(=\left[a^2-\left(b-c\right)^2\right].\left[\left(b+c\right)^2-a^2\right]\)

\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)

\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-3^2\right].\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

Tham khảo nhé~

2,  ( ax+by)²-(ax+by)²

=0

bố éo biết

\(A=4x^2+6x=2x\left(2x+3\right)\)

\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)

\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

a) \(x^5+4x+5=\left(x^5+x^4\right)-\left(x^4+x^3\right)+\left(x^3+x^2\right)-\left(x^2+x\right)+\left(5x+5\right)=x^4\left(x+1\right)-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+5\left(x+1\right)=\left(x^4-x^3+x^2-x+5\right)\left(x+1\right)\)

b) \(x^4+6x^3+11x^2+6x+1=\left(x^4+3x^3+x^2\right)+\left(3x^3+9x^2+3x\right)+\left(x^2+3x+1\right)=x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)=\left(x^2+3x+1\right)^2\)

c) \(64x^4+1=\left[\left(8x^2\right)^2+16x^2+1\right]-16x^2=\left(8x^2+1\right)^2-\left(4x\right)^2=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)d) \(81x^4+4=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)

Câu 1:

\(e,x^5-x^4-1=x^5-x^4+x^3-x^3+x^2-x^2+x-x-1\\ =\left(x^5-x^4-x^3\right)+\left(x^3-x^2-x\right)+\left(x^2-x-1\right)\\ =x^3\left(x^2-x-1\right)+x\left(x^2-x-1\right)+\left(x^2-x-1\right)\\ =\left(x^2-x-1\right)\left(x^3+x+1\right)\)

Câu 2:

\(a,\left(3x+ay+b\right)\left(x+cy+d\right)\\ =3x^2+3xcy+3xd+axy+acy^2+ayd+bx+bcy+bd\\ =3x^2+xy\left(3c+a\right)+x\left(b+3d\right)+y\left(ad+bc\right)+acy^2+bd\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}3c+a=-22\\b+3d=-4\end{matrix}\right.\\ad+bc=8\\\left\{{}\begin{matrix}ac=7\\bd=1\end{matrix}\right.\end{matrix}\right.\)

Xét \(bd=1\Leftrightarrow\left[{}\begin{matrix}b=1;d=1\\b=-1;d=-1\end{matrix}\right.\)

Với \(b=1;d=1\Leftrightarrow b+3d=1+3\cdot1=4\left(ktm\right)\)

Với \(b=-1;d=-1\Leftrightarrow b+3d=-1-3=-4\left(tm\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}3c+a=-22\\-a-c=8\\ac=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\c=-7\end{matrix}\right.\)

Vậy \(3x^2-22xy-4x+8y+7y^2+1=\left(3x-y-1\right)\left(x-7y-1\right)\)

Cái chỗ ngoặc nhọn mà 5 dòng á a ko thấy trong cái phần công thức nên là ghi z chứ nó có 5 dòng đó nha

câu b tương tự, lười wa 😴