Phân tích các đa thức sau thành nhân tử
a) 9(2x+3)2-4(x+1)2
b)27x3+y3/8
c) (x+y)3-(x-y)3
Giúp mk nha
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Phân tích các đa thức sau thành nhân tử
a) 9(2x+3)2-4(x+1)2
b)27x3+y3/8
c) (x+y)3-(x-y)3
Giúp mk nha
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
a) \(14\left(x-y\right)^2+21\left(y-x\right)\)
\(=14\left(x-y\right)^2-21\left(x-y\right)\)
\(=7\left(x-y\right)\left[2\left(x-y\right)-3\right]\)
\(=7\left(x-y\right)\left(2x-2y-3\right)\)
b) \(7x^5\left(y-3\right)-49x^4\left(3-y\right)^3\)
\(=7x^4\left(y-3\right)\left[x+7\left(y-3\right)^2\right]\)
\(=7x^4\left(y-3\right)\left(x+7y^2-42y+63\right)\)
c) \(\left(x^2-9\right)^2-x^2\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left(x+3\right)^2-x^2\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left[\left(x+3\right)^2-x^2\right]\)
\(=\left(x-3\right)^2\left(x^2+6x+9-x^2\right)\)
\(=3\left(x-3\right)^2\left(x+3\right)\)
d) \(\left(4x^2-1\right)^2-9\left(2x-1\right)^2\)
\(=\left(2x-1\right)^2\left(2x+1\right)^2-9\left(2x-1\right)^2\)
\(=\left(2x-1\right)^2\left[\left(2x+1\right)^2-9\right]\)
\(=\left(2x-1\right)^2\left(4x^2+4x+1-9\right)\)
\(=4\left(2x-1\right)^2\left(x^2+x-2\right)\)
\(=4\left(2x-1\right)^2\left(x-1\right)\left(x+2\right)\)
a: =64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2+y^2-4xy)(8x^2+y^2+4xy)
b: =x^8+2x^4+1-x^4
=(x^4+1)^2-x^4
=(x^4-x^2+1)(x^4+x^2+1)
=(x^4-x^2+1)(x^4+2x^2+1-x^2)
=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)
c: =(x+1)(x^2-x+1)+2x(x+1)
=(x+1)(x^2-x+1+2x)
=(x+1)(x^2+x+1)
d: =(x^2-1)(x^2+1)-2x(x^2-1)
=(x^2-1)(x^2-2x+1)
=(x-1)^2*(x-1)(x+1)
=(x+1)(x-1)^3
Bài 1 :
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(a=x^2+6x-7\)
\(A=a\left(a-9\right)+8\)
\(A=a^2-9a+8\)
\(A=a^2-8a-a+8\)
\(A=a\left(a-8\right)-\left(a-8\right)\)
\(A=\left(a-8\right)\left(a-1\right)\)
Thay a vào là xong bạn :)
a) \(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
b) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+y^3\right)+\left(3x^2+3xy^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+3xy-1\right)\)
\(=\left(x+y\right)\left(x^2+y^2+2xy-1\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
Câu b sorry nha
a, A =x4 - ( x2 - 2x + 1)
= x4 - ( x - 1)2
= (x2 - x + 1) ( x2 + x - 1)
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.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)
=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)
a) 9 . ( 2x + 3 )2 - 4 . ( x + 1 )2
= 32 . ( 2x + 3 )2 - 22 . ( x + 1 )
= [ 3 . ( 2x + 3 ) ] 2 - [ 2 . ( x + 1 ) ]
= ( 6x + 9 )2 - ( 2x + 2 )2
= [ ( 6x + 9 ) - ( 2x + 2 ) ] . [ ( 6x + 9 ) + ( 2x + 2 ) ]
= ( 6x + 9 - 2x - 2 ) . ( 6x + 9 + 2x + 2 )
= ( 4x + 7 ) . ( 8x + 11 )
b) 27x3 + \(\dfrac{y^3}{8}\) = ( 3x )3 + \(\left(\dfrac{y}{2}\right)^{^{ }3}\)
= ( 3x + \(\dfrac{y}{2}\) ) . ( 9x2 - \(\dfrac{3}{2}\)xy + \(\dfrac{y^2}{4}\) )
c) ( x + y )3 - ( x - y )3
= [ ( x + y ) - ( x - y )] . [ ( x + y )2 + ( x + y ) . ( x - y ) + ( x - y )2
= ( x + y - x + y ) . [ ( x2 + 2xy + y2 ) ] + ( x2 - y2 ) + ( x2 - 2xy + y2 ) ]
= 2y . ( x2 + 2xy + y2 + x2 - y2 + x2 - 2xy + y2 )
= 2y . ( 3x2 + y2 )
= 6y . ( x2 + y2 )
bỏ 6y . (x2 - y2 ) đi