x^4 + 4y^4
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\(a,=x\left(x^3+3x^2-6x-8\right)\\ =x\left(x^3+4x^2-x^2-4x-2x-8\right)\\ =x\left(x+4\right)\left(x^2-x-2\right)\\ =x\left(x+4\right)\left(x-2\right)\left(x+1\right)\)
\(b,=x^4+36x^2+324-36x^2\\ =\left(x^2+18\right)^2-36x^2\\ =\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)
\(c,=xy\left(x^3y^3+xy+2\right)\)
1.
x4y4+4=[(x2y2)2+2.x2y2.2+22]-4x2y2
=(x2y2+2)2-(2xy)2
bạn tính nốt đi, câu 2, 4, 6 tương tự
câu 4 khá dài bạn lấy số đấy chia cho (x+1) ra nháp rồi tính ngược lại sẽ ra
1: \(=x^4y^4+4+4x^2y^2-4x^2y^2\)
\(=\left(x^2y^2+2\right)^2-4x^2y^2\)
\(=\left(x^2y^2+2xy+2\right)\left(x^2y^2-2xy+2\right)\)
2: \(=x^4y^4+16x^2y^2+64-16x^2y^2\)
\(=\left(x^2y^2+8\right)^2-16x^2y^2\)
\(=\left(x^2y^2+8-4xy\right)\left(x^2y^2+8+4xy\right)\)
3: \(=x^4+4x^2+4-x^2\)
\(=\left(x^2+2\right)^2-x^2\)
\(=\left(x^2+x+2\right)\left(x^2-x+2\right)\)
4: \(=4x^4y^4+1+4x^2y^2-4x^2y^2\)
\(=\left(2x^2y^2+1\right)^2-\left(2xy\right)^2\)
\(=\left(2x^2y^2+1-2xy\right)\left(2x^2y^2+1+2xy\right)\)
6: \(=x^4+4y^4+4x^2y^2-4x^2y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2+2xy\right)\left(x^2+2y^2-2xy\right)\)
a)x3-6x2+9x=x(x2-6x+9)=x(x-3)2
b)x2-2x-4y2-4y=(x2-2x+1)-(4y2+4y+1)=(x-1)2-(2y+1)2=(x-1-2y-1)(x-1+2y+1)=(x-2y-2)(x+2y)
c)x2-x+xy-y=x(x-1)+y(x-1)=(x-1)(x+y)
d)3x2-6xy-75+3y2=3[(x2-2xy+y2)-25]=3[(x-y)2-52]=3(x-y-5)(x-y+5)
e)2x2-5x-7=(2x2+2x)-(7x+7)=2x(x+1)-7(x+1)=(x+1)(2x-7)
f)x4+36=x4+12x2+36-12x2=(x2+6)2-12x2=(x2-\(\sqrt{12}x\)+6)(x2+\(\sqrt{12}x\)+6)
h)x4+4y4=x4+4x2y2+4y2-4x2y2=(x2+2y2)-4x2y2=(x2+2y2-2xy)(x2+2y2+2xy)
x4y4 + 4
= x4y4 + 4x2y2 + 4 - 4x2y2
= (x2y2 + 2)2 - (2xy)2
= (x2y2 - 2xy + 2)(x2y2 + 2xy + 2)
x4y4 + 64
= x4y4 + 16x2y2 + 64 - 16x2y2
= (x2y2 + 8)2 - (4xy)2
= (x2y2 - 4xy + 8)(x2y2 + 4xy + 8)
x5 + x + 1
= x5 - x2 + x2 + x + 1
= x2(x3 - 1) + (x2 + x + 1)
= x2(x - 1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)[x2(x - 1) + 1]
a) \(x^2-10x+4y^2-4y+26=0\)
\(\Leftrightarrow\left(x^2-10x+25\right)+\left(4y^2-4y+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)^2+\left(2y-1\right)^2=0\)
Mà \(\Leftrightarrow\left(x-5\right)^2+\left(2y-1\right)^2\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}x-5=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{1}{2}\end{cases}}\)
x4+4y4=x4+4x2y2+4y4-4x2y2
=(x2+2y2)-4x2y2
=(x2+2y2-2xy)(x2+2y2+2xy)