phân tích đa thức thành nhân tử
\(4x^4-37x^2+9\)
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a) \(x^3\left(x^2-7\right)^2-36x=x\left[\left(x^3-7x\right)^2-6^2\right]\)
\(=x\left(x^3-7x-6\right)\left(x^3-7x+6\right)\)
\(x\left[\left(x-3\right)\left(x+1\right)\left(x+2\right)\right].\left[\left(x+3\right)\left(x-2\right)\left(x-1\right)\right]\)
\(=\left(x-3\right)\left(x-2\right)\left(x-1\right).x.\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b) Không pt được.
c) Không pt được.
4x4-37x2+9
=4x4-12x2+9-25
=(2x2-3)2-25
=(2x2-3-5)(2x2-3+5)
=(2x2-8)(2x2+2)
=2.(x2-4).2.(x2+1)
=4(x-2)(x+2)(x2+1)
x^8+x^4+1
=x8+2x4+1-x4
=(x4+1)2-x4
=(x4-x2+1)(x4+x2+1)
=(x4-x2+1)(x4+2x2+1-x2)
=(x4-x2+1)[(x2+1)2-x2]
=(x4-x2+1)(x2-x+1)(x2+x+1)
9x2-2015x+2006
= 9x2-9x-2006x+2006
= (9x2-9x)-(2006x-2006)
= 9x(x-1)-2006(x-1)
= (x-1) (9x-2006)
Chúc học tốt nhé!
a) Hình như phân tích không được
b) \(2x^4+5x^3+13x^2+25x+15\)
\(=x^3+1+2x^4+2x^3+13x^2+13x+12x+12+2+2x^3\)
\(=\left(x^3+1\right)+\left(2x^4+2x^3\right)+\left(13x^2+13x\right)+\left(12x+12\right)+2\left(1+x^3\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x^3\left(x+1\right)+13x\left(x+1\right)+12\left(x+1\right)+2\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x^3+13x+12+2x^2-2x+2\right)\)
\(=\left(x+1\right)\left(3x^2+10x+15+2x^3\right)\)
\(=\left(x+1\right)\left[x^2\left(2x+3\right)+5\left(2x+3\right)\right]\)
\(=\left(x+1\right)\left(x^2+5\right)\left(2x+3\right)\)
=\(\left(3x^2-4x-13-4x^2+9\right)\left(3x^2-4x-13+4x^2-9\right)-\left(x+2\right)^4\)
=\(\left(-x^2-4x-4\right)\left(7x^2-4x-22\right)-\)\(\left(x+2\right)^{^{ }2.2}\)
=\(-\left(x+2\right)^2\left(7x^2-4x-22\right)-\left(x+2\right)^2\left(x+2\right)^2\)
=\(-\left(x+2\right)^2\)\(\left(7x^2-4x-22-x^2-4x-4\right)\)
\(-\left(x+2\right)^2\)(\(6x^2-8x-26\))
\(4x^4+4x^2+1=\left(2x^2+1\right)^2\)
\(9x^4-6x^2+1=\left(3x^2-1\right)^2\)
\(\dfrac{x^2}{9}-\dfrac{2}{3}x+1=\left(\dfrac{x}{3}+1\right)^2\)
\(x^2-25=\left(x-5\right)\left(x+5\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(4x^4-37x^2+9=4x^4-36x^2-x^2+9=4x^2\left(x^2-9\right)-\left(x^2-9\right).\)
\(=\left(x^2-9\right)\left(4x^2-1\right)=\left(x-3\right)\left(x+3\right)\left(2x-1\right)\left(2x+1\right)\)
Đặt t = x2
đa thức trở thành 4t2 - 37t + 9
= 4t2 - t - 36t + 9
= ( 4t2 - 36t ) - ( t - 9 )
= 4t( t - 9 ) - ( t - 9 )
= ( t - 9 )( 4t - 1 )
= ( x2 - 9 )( 4x2 - 1 )
= ( x - 3 )( x + 3 )( 2x - 1 )( 2x + 1 )