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\(\left(x^2-x-3\right)\left(x^2-x-4\right)-12\)
\(=x^4-x^3-4x^2-x^3+x^2+4x-3x^2+3x+12-12=x^4-2x^3-6x^2+7x\)
\(=x.\left(x^3-2x^2-6x+7\right)=x.\left(x^3-x^2-x^2+x-7x+7\right)\)
\(=x.\left[x^2.\left(x-1\right)-x.\left(x-1\right)-7.\left(x-1\right)\right]=x\left(x-1\right)\left(x^2-x-7\right)\)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
Ta có : 3(x4 + x2 + 1) - (x2 + x + 1)2
= 3(x4 + x2 + 1) - x4 - x² - 1 - 2x3 - 2x - 2x2
= 3(x4 + x2 + 1) - (x4 + x² + 1) - 2(x3 + x + x2)
= 2(x4 + x2 + 1) - 2(x3 + x + x2)
= 2(x4 + x2 + 1 - x3 - x - x2)
= 2(x4 - x3 - x + 1)
Đơn giản thôi :]>
Sau khi phân tích thì P(x) có dạng ( x2 + dx + 2 )( x2 + ax - 2 )
P(x) = x4 - x3 - 2x - 4 = ( x2 + dx + 2 )( x2 + ax - 2 )
⇔ x4 - x3 - 2x - 4 = x4 + ax3 - 2x2 + dx3 + adx2 - 2dx + 2x2 + 2ax - 4
⇔ x4 - x3 - 2x - 4 = x4 + ( a + d )x3 + adx2 + ( 2a - 2d )x - 4
Đồng nhất hệ số ta được :
\(\hept{\begin{cases}a+d=-1\\ad=0\\2a-2d=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-1\\d=0\end{cases}}\)
( x2 + dx + 2 )( x2 + ax - 2 )
= ( x2 + 2 )( x2 - x - 2 )
= ( x2 + 2 )( x2 - 2x + x - 2 )
= ( x2 + 2 )[ x( x - 2 ) + ( x - 2 ) ]
= ( x2 + 2 )( x - 2 )( x + 1 )
=> P(x) = x4 - x3 - 2x - 4 = ( x2 + 2 )( x - 2 )( x + 1 )
\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)
x^3-2x-4
=x^3-4x+2x-4
=x(x^2-4)+2(x-2)
=x(x-2)(x+2)+2(x-2)
=(x^2+2x+2)(x-2)
Nhớ cho mình nha!!!
Dễ thôi
\(x^3-x^2-4\)
\(=\left(x^3-8\right)-\left(x^2-4\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x-2\right)\left(x^2+x+2\right)\)