Phân tích đa thức thành nhân tử :
\(x^4+2004x^2+2003x+2004\)
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TT
2
7 tháng 10 2018
\(x^4+2004x^2+2003x+2004\)
\(=x^4-x+2004x^2+2004x+2004\)
\(=x\left(x^3-1\right)+2004\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2004\right)\)
7 tháng 10 2018
\(x^4+2004x^2+2003x+2004\)
\(=x^4+2004x^2+2004x-x+2004\)
\(=\left(x^4-x\right)+2004\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)+2004\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2004\right)\)
HD
1
4 tháng 2 2016
x^4+2005x^2+2004x+2005
=x^4-x+2005x^2+2005x+2005
=x(x^3-1)+2005(x^2+x+1)
=x(x-1)(x^2+x+1)+2005(x^2+x+1)
=(x^2+x+1)(x^2-x+2005)
PT
1
MT
9 tháng 7 2015
x4 + 2005x2 + 2004x + 2005
=x4-x+2005x2+2005x+2005
=x(x3-1)+2005.(x2+x+1)
=x(x-1)(x2+x+1)+2005.(x2+x+1)
=(x2+x+1)[x(x-1)+2005]
=(x2+x+1)(x2-x+2005)
JI
0
JI
0
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
LT
1
MT
9 tháng 7 2015
x4 + 2005x2 + 2004x + 2005
=x4+2005x2+2005x-x+2005
=x4-x+2005x2+2005x+2005
=x(x3-1)+2005.(x2+x+1)
=x(x-1)(x2+x+1)+2005.(x2+x+1)
=(x2+x+1)[x(x-1)+2005]
=(x2+x+1)(x2-x+2005)
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(x^4+2004x^2+2003x+2004\)
\(=x^4+2004x^2+2004x-x+2004\)
\(=\left(x^4-x\right)+2004\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)+2004\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2004\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2004\right)\)
tìm cặp số a,b thoả mãn điều kiện; 3b/a^2-4=1-125a-3b/6a+13=1-125a