Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)
2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)
3) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)
\(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)
5) \(x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)=\left(x-1\right)^2\left(x+1\right)\)
6) \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
1) \(x^4+4=\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
2) \(a^4+64=\left(a^2+8\right)-16a^2=\left(a^2+4a+8\right)\left(a^2-4a+8\right)\)
3) \(x^5+x+1\)
\(=\left(x^5-x^4+x^2\right)+\left(x^4-x^3+x\right)+\left(x^3-x^2+1\right)\)
\(=x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
4) \(x^5+x-1\)
\(=\left(x^5+x^4-x^2\right)-\left(x^4+x^3-x\right)+\left(x^3+x^2-1\right)\)
\(=x^2\left(x^3+x^2-1\right)-x\left(x^3+x^2-1\right)+\left(x^3+x^2-1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)
bài này 1h rùi,chắc chờ tui ngủ dậy làm;
= (x+y)3 - (x+y) + xy(x+y) =
= (x+y)((x+y)2 -1 +xy)) = (x+y)(x2 +3xy +y2 -1)
a) \(7\left(3x-2\right)+y\left(3x-2\right)=\left(3x-2\right)\left(7+y\right)\)
b) \(x\left(y-x\right)-3\left(x-y\right)=x\left(y-x\right)+3\left(y-x\right)=\left(y-x\right)\left(x+3\right)\)
c) \(x^2-6xy+9y^2=\left(x-3y\right)^2\)
\(a,=\left(2y^2-1\right)\left(2y^2+1\right)\\ b,=\left(x+y\right)^2-9=\left(x+y+3\right)\left(x+y-3\right)\)
Lời giải:
a. $4y^4-1=(2y^2)^2-1^2=(2y^2-1)(2y^2+1)$
b. $x^2+2xy-9+y^2=(x^2+2xy+y^2)-9$
$=(x+y)^2-3^2=(x+y-3)(x+y+3)$
\(-3xy^2+x^2y^2-5x^2y\)
\(=-xy\left(3y+xy-5x\right)\)
\(x\left(y-1\right)+3\left(y^3+2y+1\right)\)
\(=3y^3+6y+3+xy-x\)
Xem lại nhé ko phân tích được
\(12xy^2-12xy+3x\)
\(=3x\left(4y^2-4y+1\right)\)
\(=3x\left(2y-1\right)^2\)
\(10x^2\left(x+y\right)-5\left(2x+2y\right)y^2\)
\(=10x^2\left(x+y\right)-10\left(x+y\right)y^2\)
\(=10\left(x+y\right)\left(x-y\right)\left(x+y\right)\)
\(=10\left(x+y\right)^2\left(x-y\right)\)
1) đặt 2x+1 = a => \(a^4-3a^2+2=\left(a^2-1\right)\left(a^2-2\right)=\)\(\left(a-1\right)\left(a+1\right)\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\)
=(2x+1-1)(2x+1+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\)) = 4x(x+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\))
2) =(x2-x)(x2-x-2)-3
đặt x2-x = b => b(b-2)-3 = b2-2b-3 = (b+1)(b-3) = (x2-x+1)(x2-x-3)
3) đặt x2+2x-1 = c => c2-3xc+2x2 = (c-x)(c-2x) = (x2+2x-1-x)(x2+2x-1-2x) = (x2+x-1)(x2-1) = (x2+x-1)(x-1)(x+1)
tìm x
x3-8 +(x-2)(x+1)=0 <=> (x-2)(x2+2x+4)+(x-2)(x+1)=0 <=>(x-2)(x2+2x+4+x+1)=0 <=> x=2 (vì x2+3x+5= (x+\(\frac{3}{2}\))2 +\(\frac{11}{4}\)>0)
vậy x=2
2) \(x\left(x-1\right)\left(x+1\right)\left(x-2\right)-3\)
\(=\left(x^2-x\right)\left(x^2-x-2\right)-3\)(1)
Đặt \(x^2-x=t\)
\(\Rightarrow\left(1\right)=t\left(t-2\right)-3=t^2-2t+1-4\)
\(=\left(t-1\right)^2-4\)
\(=\left(t+3\right)\left(t-5\right)\)
Thay \(x^2-x=t\), ta được:
\(BTDNT=\left(x^2-x+3\right)\left(x^2-x-5\right)\)
