Phân tích các đa thức sau thành nhân tử:
a ) x 3 + 4 x 2 – 2 x – 8
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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)
b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)
b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)
=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)
Trả lời:
a, x4 + 3x3 + x2 + 3x
= ( x4 + 3x3 ) + ( x2 + 3x )
= x3 ( x + 3 ) + x ( x + 3 )
= ( x3 + x ) ( x + 3 )
= x ( x2 + 1 ) ( x + 3 )
b, Sửa đề: x4 - x2 + 8x - 8
= ( x4 - x2 ) + ( 8x - 8 )
= x2 ( x2 - 1 ) + 8 ( x - 1 )
= x2 ( x - 1 ) ( x + 1 ) + 8 ( x - 1 )
= ( x - 1 ) [ x2 ( x + 1 ) + 8 ]
= ( x - 1 ) ( x3 + x2 + 8 )
Ta có : 5x(x - 2y) + 2(2y - x)2
= 5x(x - 2y) + 2(x - 2y)2 (vì (2y - x)2 = (x - 2y)2 )
= (x - 2y)[5x + 2(x - 2y)]
= (x - 2y)(5x + 2x - 4y)
= (x - 2y)(7x - 4y)
b) 7x(y - 4)2 - (4 - y)3
= 7x(y - 4)2 - (4 - y)2(4 - y)
= 7x(y - 4)2 - (y - 4)2(4 - y)
= (y - 4)2(7x - 4 + y)
c) (4x - 8)(x2 + 6) - (4x - 8)(x + 7) + 9(8 - 4x)
= (4x - 8)(x2 + 6) - (4x - 8)(x + 7) - 9(4x - 8)
= (4x - 8)(x2 + 6 - x - 7 - 9)
= 2(x - 4)(x2 - x - 10)
x4 + x3 + 2x2 + x + 1
= (x4 + 2x2 + 1) + (x3 + x)
= (x2 + 1)2 + x (x2 + 1)
= (x2 + 1) ( x2 + 1 + x)
= (x2 + 1) (x + 1)2
a) x3 + 4x2 – 2x – 8
= (x3 + 4x2) - (2x + 8)
= x2(x + 4) - 2(x + 4)
= (x + 4)(x2 - 2)
= (x + 4)(x + √2)(x - √2)