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\(x^3-5x^2-14x\)
\(=x^3+2x^2-7x^2-14x\)
\(=x^2\left(x+2\right)-7x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-7x\right)\)
\(=x\left(x+2\right)\left(x-7\right)\)
\(x^3-7x-6\)
\(=x^3+x^2-x^2-x-6x-6\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x^2+2x-3x-6\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x-3\right)\)
\(x^3-19x-30\)
\(=x^3-5x^2+5x^2-25x+6x-30\)
\(=x^2\left(x-5\right)+5x\left(x-5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+3\right)\left(x+2\right)\)
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a) \(4x^4+4x^3+5x^2+2x+1\)
= \(x^2\left(4x^2+4x+5+\frac{4}{x}+\frac{1}{x^2}\right)\)
=\(x^2\left[\left(4x^2+\frac{1}{x^2}\right)+2\left(2x+\frac{1}{x}\right)+5\right]\)(1)
Đặt \(2x+\frac{1}{x}=a\)thì \(\left(2x+\frac{1}{x}\right)^2=a^2\)\(\Rightarrow4x^2+\frac{1}{x^2}=a^2-4\)
Thay vào (1), ta có:
\(x^2\left(a^2-4+2a+5\right)\)
=\(x^2\left(a^2+2a+1\right)\)
=\(x^2\left(a+1\right)^2\)
=\(\left[x\left(a+1\right)\right]^2\)
=\(\left[x\left(2x+\frac{1}{x}+1\right)\right]^2\)
=\(\left(2x^2+1+x\right)^2\)
\(=\left(2x^2+x+1\right)^2\)
a) Đặt f(x) = 4x4 + 4x3 + 5x2 + 2x + 1
Sau khi phân tích thì đa thức có dạng ( 2x2 + ax + 1 )( 2x2 + bx + 1 )
=> f(x) = ( 2x2 + ax + 1 )( 2x2 + bx + 1 )
<=> f(x) = 4x4 + 2bx3 + 2x2 + 2ax3 + abx2 + ax + 2x2 + bx + 1
<=> f(x) = 4x4 + ( a + b )2x3 + ( ab + 4 )x2 + ( a + b )x + 1
Đồng nhất hệ số ta có : \(\hept{\begin{cases}a+b=2\\ab=1\end{cases}\Leftrightarrow}a=b=1\)
Vậy f(x) = 4x4 + 4x3 + 5x2 + 2x + 1 = ( 2x2 + x + 1 )2
b) 3x4 + 11x3 - 7x2 - 2x + 1
= 3x4 - x3 + 12x3 - 4x2 - 3x2 + x - 3x + 1
= x3( 3x - 1 ) + 4x2( 3x - 1 ) - x( 3x - 1 ) - ( 3x - 1 )
= ( 3x - 1 )( x3 + 4x2 - x - 1 )
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x3 + 7x - 6=x2 . x + 7x - 22 + 2 = (x2 - 22) + (x+7x)+2=(x-2) . (x+2) + 8x + 2
x3 - 5x + 8x - 4=x2 . x -5x + 8x -22 = (x2 - 22) . (x -5x + 8x )=(x-2) . (x+2) . 4x
x3 - 9x2 + 6x + 16=x2 . x - 9x2 + 6x + 16 = (x2 - 9x2) . (x+6x) + 16=(x-9x) . (x+9x) . 7x + 16
k mk nha
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Ta có : \(4x^2-3x-1\)
\(=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(4x+1\right)\)
Ta có : \(x^2-7x+12\)
\(=x^2-3x-4x+12\)
\(=x\left(x-3\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
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b) 2x^2 + 7x - 15
2x^2 + 10x - 3x -15
2x(x+5) - 3(x+5)
(x+5)(2x-3)
a, 5x^3y - 10x^2y^2 + 5xy^3 = 5xy. ( x^2 - 2xy + y^2) = 5xy.( x-y)^2
b, 2x^2 + 7x -15 = 2x^2 + 10X - 3x -15
= 2x( x+5) - 3( x+5)
= ( 2x-3) (x+5)
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\(x^4-5x^3+7x^2-6\)
\(=x^4-3x^3+3x^2-2x^3+6x^2-6x-2x^2+6x-6\)
\(=x^2\left(x^2-3x+3\right)-2x\left(x^2-3x+3\right)-2\left(x^2-3x+3\right)\)
\(=\left(x^2-3x+3\right)\left(x^2-2x-2\right)\)
\(\left(x^2-x+6\right)^2+\left(x-3\right)^2\)
\(=x^4+x^2+36-2x^3-12x+12x^2+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4-2x^3+5x^2+9x^2-18x+45\)
\(=x^2\left(x^2-2x+5\right)+9\left(x^2-2x+5\right)=\left(x^2-2x+5\right)\left(x^2+9\right)\)
Bài này hay và khó đấy. Chúc bạn học tốt.
Cái bạn ghi trong đề là phương trình, còn $5x^2-7x-3$ mới là đa thức. Bạn xem lại.
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