a) \(x^4+324\)

\(=\left(x^2\right)^2+18^2+2.x^2.18-36x^2\)

\(=\left(x^2-18\right)^2-\left(6x\right)^2\)

\(=\left(x^2+18+6x\right)\left(x^2+18-6x\right)\)

b) \(64a^2+b^8\)

\(=\left(8a^2\right)^2+\left(b^4\right)^2+2.8a^2.4b^4-16a^2b^4\)

\(=\left(8a^2+b^4\right)^2-\left(ab^2\right)^2\)

\(=\left(8a^2+b^4+4ab^2\right)\left(8a^2+b^4-4ab^2\right)\)

\(a.\)

\(x^4+324\)

\(=\left(x^2\right)^2+18^2\)

\(=\left(x^2+18\right)\left(x^2_{ }-18\right)\)

\(b.\)

\(64a^2+b^8\)

\(=\left(8a^2\right)+\left(b^3\right)^2\)

\(\left(8a-b^3\right)\left(8a+b^3\right)\)

\(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

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

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

x⁴+4

=x⁴+4x²+4-4x²

=(x²+2)²-4x²

=(x²-2x+2)(x²+2x+2)

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

Ta có : \(x^4 -4 = (x^2)^2 -2^2 = ( x^2 -2)(x^2+2)\)

= (x^2-2)(x^2+2)

a, 4x⁴+1 

=(2x)²+1

=(2x+1)(2x-1)

b,c tách làm bình phương rồi làm tương tự

Ko có hằng đẳng thức đó bạn ơi../

thảo nguyên xanh xàm

(2x)^2+1 chứ cs phải (2x)^2-1 mô mà =(2x+1)(2x-1)

x⁶+3x⁴y²-8x³y³+3x²y⁴+y⁶= x⁶+3x⁴y²+3x²y⁴+y⁶-8x³y³=(x²+y²)³-(2xy)³

= (x²+y²-2xy)[(x²+y²)²+2xy(x²+y²)+(2xy)²]= (x-y)²(x⁴+6x²y²+y⁴+2x³y+2xy³)

(x²+y²-5)²-4x²y²-16xy-16=(x²+y²-5)²-(4x²y²+16xy+16)=(x²+y²-5)²-(2xy+4)²

=(x²+y²-5+2xy+4)(x²+y²-5-2xy-4)=(x²+2xy+y²-1)(x²-2xy+y²-9)=[(x+y)²-1][(x-y)²-3²]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)

x⁴+324=x⁴+36x²+324-36x²=(x²+18)²-(6x)²=(x²+18-6x)(x²+18+6x)

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

#)Giải :

\(x^3-2x-4\)

\(=x^3+2x^2-2x^2+2x-4x-4\)

\(=x^3+2x^2+2x-2x^2-4x-4\)

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

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

\(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

\(=x^3-2x^2+2x^2-4x+2x-4\)

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

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

cảm ơn nha!

để mik kb cho.mik cũng đang buồn nè. nhớ đồng ý nhaaaaa

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)