\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\)

\(=4\left(x+5\right)\left(x+10\right)\left(x+12\right)\left(x+6\right)-3x^2\)

\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

\(=4\left(x+60\right)^2+132x\left(x+60\right)+1088x^2-3x^2\)

\(=4\left(x+60\right)^2+132x\left(x+60\right)+1085x^2\)

\(=4\left(x+60\right)^2+62x\left(x+60\right)+70x\left(x+60\right)+1085x^2\)

\(=2\left(x+60\right)\left[2\left(x+60\right)+31x\right]+35x\left[2\left(x+60\right)+31x\right]\)

\(=\left(33x+120\right)\left(2x+120+35x\right)\)

\(=3\left(11x+40\right)\left(37x+120\right)\)

e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

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 😴

a)3x²+4x-9x-12=0

=>(3x²+4x)-(9x+12)=0

=> x(3x+4)-3(3x+4)=0

=> (x-3)(3x+4)=0  =>x-3=0 hoặc 3x+4=0

=>tự tính

b)7x²-9x+2=0

=>7x²-7x-2x+2=0

=>(7x²-7x)-(2x-2)=0

=>7x(x-1)-2(x-1)=0

=>(7x-2)(x-1)=0

=>như câu a

bạn chỉ biết làm 2 câu thôi