Phân tích đa thức thành nhân tử:
\(x^2-x-y^2-y\) \(a^3-a^2x-ay+xy\)
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1) x2 - x - y2 - y = (x - y)(x + y) - (x + y) = (x - y - 1)(x + y)
2. x2 - 2xy + y2 - z2 = (x - y)2 - z2 = (x - y - z)(x - y + z)
3. 5x - 5y + ax - ay = 5(x - y) + a(x - y) = (a + 5)(x - y)
4. a3 - a2x - ay + xy = a2(a - x) - y(a - x) = (a2 - y)(a - x)
5. 4x2 - y2 + 4x + 1 = (2x + 1)2 - y2 = (2x + 1 - y)(2x + y + 1)
6. x3 - x + y3 - y = (x + y)(x2 - xy + y2) - (x + y) = (x + y)(x2 - xy + y2 - 1)
Trả lời:
1, x2 - x - y2 - y
= ( x2 - y2 ) - ( x + y )
= ( x - y ) ( x + y ) - ( x + y )
= ( x + y ) ( x - y - 1 )
2, x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - x2
= ( x - y - z ) ( x - y + z )
3, 5x - 5y + ax - ay
= ( 5x + ax ) - ( 5y + ay )
= x ( 5 + a ) - y ( 5 + a )
= ( 5 + a ) ( x - y )
= ( 5 + a ) ( x - y )
4, a3 - a2x - ay + xy
= ( a3 - a2x ) - ( ay - xy )
= a2 ( a - x ) - y ( a - x )
= ( a - x ) ( a2 - y )
5, 4x2 - y2 + 4x + 1
= ( 4x2 + 4x + 1 ) - y2
= ( 2x + 1 )2 - y2
= ( 2x + 1 - y ) ( 2x + 1 + y )
6, x3 - x + y3 - y
= ( x3 + y3 ) - ( x + y )
= ( x + y ) ( x2 - xy + y ) - ( x + y )
= ( x + y ) ( x2 - xy + y - 1 )
x2 - x - y2 - y
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
***
9x2 + y2 - 16z2 + 6xy
= (3x + y)2 - (4z)2
= (3x + y - 4z)(3x + y + 4z)
***
a3 - a2x - ay + xy
= a2(a - x) - y(a - x)
= (a - x)(a2 - y)
***
2x2 - 8y2 + 3x + 6y
= 2(x2 - 4y2) + 3(x + 2y)
= 2(x - 2y)(x + 2y) + 3(x + 2y)
= (x + 2y)(2x - 4y + 3)
***
xy(x + y) + yz(y + z) + xz(x + z) + 2xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + z)
= y(x + y + z)(x + z) + xz(x + z)
= (x + z)(xy + y2 + yz + xz)
= (x + z)[y(x + y) + z(x + y)]
= (x + z)(x + y)(y + z)
a Đề sai: )
b
\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)
c
\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)
d
\(x^4+2x^3+x^2\\ =x^4+x^3+x^3+x^2\\ =x^3\left(x+1\right)+x^2\left(x+1\right)\\ =\left(x^3+x^2\right)\left(x+1\right)\)
e
\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)
c: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
d: =x^2(x^2+2x+1)
=x^2(x+1)^2
e: =5(x^2-2xy+y^2-z^2)
=5[(x-y)^2-z^2]
=5(x-y-z)(x-y+z)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(a^3-a^2x-ay+xy\)
\(=a^2\left(a-x\right)-y\left(a-x\right)\)
\(=\left(a-x\right)\left(a^2-y\right)\)
\(4x^2-y^2+4x+1\)
\(=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2=\left(2x-y+1\right)\left(2x+y+1\right)\)
\(x^3-x+y^3-y\)
\(=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
a)a3 - a2x - ay +xy
=(a3 - a2x) - (ay - xy)
=a2(a-x) - y(a-x)
=(a-x).(a2 - y)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right).\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right).\left(x-y-1\right)\)
\(a^3-a^2x-ay+xy\)
\(=\left(a^3-a^2x\right)-\left(ay-xy\right)\)
\(=a^2.\left(a-x\right)-y.\left(a-x\right)\)
\(=\left(a-x\right).\left(a^2-y\right)\)
Ủng hộ mk nha ^_-
xét x2-x-y2-y
=x^2-y^2-(x+y)
=(x+y)(x-y)+(x+y)
=(x+y)(x-y+1)
Câu tiep theo cau nhom hang tu thu nhat voi thu 3,thu 2 voi thu4