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a, 4x2 - 4x - 3
=4x2-2x+6x-3
=2x(2x-1)+3(2x-1)
=(2x+3)(2x-1)
b, x3 - x2 - 4
= x3-x2+0x-4
= x3-2x2+x2-2x+2x-4
= (x3-2x2)+(x2-2x)+(2x-4)
= x2(x-2)+x(x-2)+2(x-2)
=(x-2)(x2+x+2)
c, 64x4+y4
=64x4+16x2y2+y4-16x2y2
= (8x2+y2)2-16x2y2
= (8x2+y2-4xy)(8x2+y2+4xy)
\(64x^4+y^4\)
\(=\left(8x^2\right)^2+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
Bài 1:
a) \(x^2-2xy-25+y^2\) (Sửa đề)
\(=x^2-2xy+y^2-25\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
Vậy ...
b) \(x\left(x-1\right)+y\left(1-x\right)\)
\(=x\left(x-1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
Vậy ...
c) \(7x+7y-\left(x+y\right)\) (Sửa đề)
\(=7\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(7-1\right)\)
\(=6\left(x+y\right)\)
Vậy ...
d) \(x^4+y^4\)
\(=\left(x^2\right)^2+\left(y^2\right)^2\)
\(=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\left(x^2+y^2-\sqrt{2}xy\right)\left(x^2+y^2+\sqrt{2}xy\right)\)
Vậy ...
\(a^2+2ab+b^2-x^2-2xy-y^2=\left(a+b\right)^2-\left(x+y\right)^2=\left(a+b+x+y\right)\left(a+b-x-y\right)\)\(x+2y-xy-2=x-xy+2y-2=x\left(1-y\right)-2\left(1-y\right)=\left(x-2\right)\left(1-y\right)\)
\(x^5+x^4+1=x^5-x^2+x^4-x+\left(x^2+x+1\right)=x^2\left(x^3-1\right)+x\left(x^3-1\right)+\left(x^2+x+1\right)=x^2\left(x-1\right)\left(x^2+x+1\right)+x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^3-x+1\right)\)\(64x^4+y^4=64x^4+16x^2y^2+y^4-16x^2y^2=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
\(x^3+8x^2+17x+10=\left(x^3+2x^2\right)+\left(6x^2+12x\right)+\left(5x+10\right)=x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)=\left(x^2+6x+5\right)\left(x+2\right)=\left(x+1\right)\left(x+2\right)\left(x+5\right)\) \(4x^4+81=4x^4+36x^2+81-36x^2=\left(2x^2+9\right)^2-\left(6x\right)^2=\left(2x^2+6x+9\right)\left(2x^2-6x+9\right)\)\(abc+ab+bc+ca+a+b+c+1=\left(abc+ab\right)+\left(bc+b\right)+\left(ca+a\right)+\left(c+1\right)=ab\left(c+1\right)+b\left(c+1\right)+a\left(c+1\right)+\left(c+1\right)=\left(c+1\right)\left(ab+a+b+1\right)=\left(c+1\right)\left(a+1\right)\left(b+1\right)\)
a) x4 - 4x2 + 4x - 1
= ( x2)2 - [ ( 2x)2 - 2.2x + 1]
= ( x2)2 - ( 2x - 1)2
= ( x2 - 2x +1)( x2 + 2x - 1)
= ( x -1)2( x2 + 2x - 1)
b) 4x2 - y2 + 4x + 1
= (2x)2 + 2.2x +1 - y2
= ( 2x +1)2 - y2
= ( 2x + 1 - y)( 2x + 1 + y)
\(\text{a) }x^4-4x^2+4x-1\\ \\=x^4-\left(4x^2-4x+1\right)\\ \\ =\left(x^2\right)^2-\left(2x-1\right)^2\\ \\=\left(x^2-2x+1\right)\left(x^2+2x-1\right)\\ \\=\left(x-1\right)^2\left(x^2+2x-1\right)\)
\(\text{b) }4x^2-y^2+4x+1\\ \\=\left(4x^2+4x+1\right)-y^2\\ \\=\left(2x+1\right)^2-y^2\\ \\=\left(2x+1+y\right)\left(2x+1-y\right)\)
a)\(x^3+3xy+y^3-1\)
\(=x^3+3x^2y+3xy^2+y^3-1-3x^2y-3xy^2+3xy\)
\(=\left(x+y\right)^3-1^3-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
b) Đặt \(B=3x^2+22xy+11x+37y+7y^2+10\)
Giả sử \(B=\left(ax+by+c\right)\left(mx+ny+p\right)\)
\(=amx^2+anxy+apx+bmxy+bny^2+bpy+cmx+cny+cp\)
\(=amx^2+\left(an+bm\right)xy+\left(ap+cm\right)x+bny^2+\left(bp+cn\right)y+cp\)
Ta được hệ: \(\left\{{}\begin{matrix}am=3;an+bm=22\\ap+cm=11;bn=7\\bp+cn=37;cp=10\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=3;b=1\\c=5;m=1\\n=7;p=2\end{matrix}\right.\)
Vậy B phân tích được thành \(\left(3x+y+5\right)\left(x+7y+2\right)\).
a/ =(x+y)3-1-3xy(x+y-1)
=(x+y-1)(x2+2xy+y2+xy+1)-3xy(x+y-1)
=(x+y-1)(x2+y2+1)
mơn nha
em ko lam dc anh ehh
em chua gap bai nao nhu the nay. noi dung hon la chua den lop lam haha
a. \(\left(xy+1\right)^2-\left(x+y\right)^2=\left(xy+1+x+y\right)\left(xy+1-x-y\right)=\left(x+1\right)\left(y+1\right)\left(x-1\right)\left(y-1\right)\)
b.
thêm bớt 16x^2y^2
\(64x^4+y^4=64x^4+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)