Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(6x^4-9x^3\)
\(=3x^3\left(2x-3\right)\)
b) \(5y^{10}+15y^6\)
\(=5y^6\left(y^4+5\right)\)
c) \(9x^2y^2+15^2y-21xy^2\)
\(=9x^2y^2+225y-21xy^2\)
\(=3y\left(3x^2y+75-7xy\right)\)
d) \(x^2y^2z+xy^2z^2+x^2yz^2\)
\(=xyz\left(xy+yz+xz\right)\)
a)\(a^4+a^2+1=\left(a^2\right)^2+2a^2.1+1^2-a^2=\left(a^2+1\right)^2-a^2=\left(a^2+1+a\right)\left(a^2+1-a\right)\)
b)\(a^4+a^2-2=a^4-a^2+2a^2-2=a^2\left(a^2-1\right)+2\left(a^2-1\right)=\left(a^2+2\right)\left(a^2-1\right)\)
c)\(x^4+4x^2-5=x^4-x^2+5x^2-5=x^2\left(x^2-1\right)+5\left(x^2-1\right)=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)
d)\(\left(x+2\right)\left(x^2-2x-6\right)=x^3-2x^2-6x+2x^2-4x-12=x^3-10x-12\)
\(\Rightarrow x^3-10x-12=\left(x+2\right)\left(x^2-2x-6\right)\)
e)\(6x^3-17x^2+14x-3\)
Ta có: \(\left(ax^2+bx+c\right)\left(dx+e\right)\)
\(=adx^3+aex^2+bdx^2+bex+cdx+ce\)
\(=adx^3+\left(ae+bd\right)x^2+\left(be+cd\right)x+ce\)
Do đó:\(\left\{{}\begin{matrix}ad=6\\ae+bd=-17\\be+cd=14\\ce=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3;b=-4\\c=1;d=2\\e=-3\end{matrix}\right.\)
Suy ra: \(6x^3-17x^2+14x-3=\left(3x^2-4x+1\right)\left(2x-3\right)\)
h)\(x^4-34x^2+225=x^4-15x^2-15x^2+225-4x^2=x^2\left(x^2-15\right)-15\left(x^2-15\right)-\left(2x\right)^2=\left(x^2-15\right)^2-\left(2x\right)^2=\left(x^2+2x-15\right)\left(x^2-2x-15\right)=\left(x^2-3x+5x-15\right)\left(x^2+5x-3x-15\right)=\left[\left(x-3\right)\left(x+5\right)\right]^2\)
\(1\hept{\begin{cases}6x^2-8x+3x-4\\2x\left(3x-4\right)+\left(3x-4\right)\\\left(3x-4\right)\left(2x+1\right)\end{cases}}\)
\(2\hept{\begin{cases}7x^2-7xy-5x+5y+6xy\\7x\left(x-y\right)-5\left(x-y\right)+\frac{6xy\left(x-y\right)}{\left(x-y\right)}\\\left(x-y\right)\left(7x-5+\frac{6xy}{\left(x-y\right)}\right)\end{cases}}\)
\(3\hept{\begin{cases}5x\left(x-y\right)-15\left(x-y\right)\\\left(x-y\right)\left(5x-15\right)\end{cases}}\)
\(4,,2x^2+x=x\left(2x+1\right)\)
\(5\hept{\begin{cases}x^3-4x-3x^2+12\\x\left(x^2-4\right)-3\left(x^2-4\right)\\\left(x+2\right)\left(x-2\right)\left(x-3\right)\end{cases}}\)
\(6\hept{\begin{cases}2x+2y+x^2-y^2\\2\left(x+y\right)+\left(x+y\right)\left(x-y\right)\\\left(x+y\right)\left(2+x-y\right)\end{cases}}\)
\(7\hept{\begin{cases}\left(x^2y-2xy\right)-\left(xy-2y\right)+\left(xy-y\right)\\xy\left(x-2\right)-y\left(x-2\right)+y\left(x-1\right)\\y\left(X-2\right)\left(x-1\right)+y\left(x-1\right)\end{cases}}\Leftrightarrow y\left(x-1\right)\left(x-2+1\right)\)
\(8\hept{\begin{cases}x\left(2-y\right)+z\left(2-y\right)\\\left(2-y\right)\left(x+1\right)\end{cases}}\)
Bài nhiều quá... nhìn mik nổi gai ốc lun...oh my god sao mà nhiều vậy nè .
Mik định giải giúp bạn nhưng bây h mik hoảng quá ... nhiều vậy chắc mik chết mất... ToT ... >.< =)))
a) \(12x^5y+24x^4y^2+12x^3y^3\)
\(=12x^3y\left(x^2+2xy+y^2\right)\)
\(=12x^3y\left(x+y\right)^2\)
b) \(x^2-2xy-4+y^2\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
g) \(12xy-12xz+3x^2y-3x^2z\)
\(=12x\left(y-z\right)+3x^2\left(y-z\right)\)
\(=3x\left(4+x\right)\left(y-z\right)\)
e) \(16x^2-9\left(x^2+2xy+y^2\right)\)
\(=\left(4x\right)^2-\left[3\left(x+y\right)\right]^2\)
\(=\left(4x-3\left(x+y\right)\right)\left(4x+3\left(x+y\right)\right)\)
\(=\left(x+y\right)\left(7x+y\right)\)
d) làm tương tự như phần g chỉ khác là phải nhóm( nhóm xen kẽ), phần f cũng vậy
a) -x2 + 2x - 1
= -( x2 - 2x + 1 )
= -( x - 1 )2
b) 12y - 36 - y2
= -( y2 - 12y + 36 )
= -( y - 6 )2
c) -x3 + 9x2 - 27x + 27
= -( x3 - 9x2 + 27x - 27 )
= -( x - 3 )3
d) x3 - 6x2 + 9x
= x( x2 - 6x + 9 )
= x( x - 3 )2
e) a3b - ab3
= ab( a2 - b2 )
= ab( a - b )( a + b )
f) a2 + 2a + 1 - b2
= a2 + ab + a - ab - b2 - b + a + b + 1
= a( a + b + 1 ) - b( a + b + 1 ) + 1( a + b + 1 )
= ( a - b + 1 )( a + b + 1 )
a)\(-x^2+2x-1\)
\(=-\left(x^2-2x+1\right)\)
\(=-\left(x-1\right)^2\)
b) \(12y-36-y^2\)
\(=-\left(y^2-12y+36\right)\)
\(=-\left(y^2-2\cdot1\cdot6+6^2\right)\)
\(=-\left(y-6\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-x^3+3x^2+6x^2-18x-9x+27\)
\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x-3\right)\left(-x^2+6x-9\right)\)
\(=\left(x-3\right)\cdot-\left(x^2-6x+9\right)\)
\(=\left(x-3\right)\cdot-\left(x^2-2\cdot x\cdot3+3^2\right)\)
\(=-\left(x-3\right)\left(x-3\right)^2\)
\(=\left(x-3\right)^3\)
d) \(x^3-6x^2+9\)
\(=x\left(x^2-6x+9\right)\)
\(=x\left(x-3\right)^2\)
e) \(a^3b-ab^3\)
\(=ab\left(a^2-b^2\right)\)
\(=ab\left(a-b\right)\left(a+b\right)\)
f) \(a^2+2a+1-b^2\)
\(=a^2+2\cdot a\cdot1+1^2-b^2\)
\(=\left(a+1\right)^2-b^2\)
\(=\left(a+1-b\right)\left(a+1+b\right)\)
a) Ta có: \(x^2+9x+20\)
\(=x^2+4x+5x+20\)
\(=x\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(x+5\right)\)
b) Ta có: \(x^2+x-12\)
\(=x^2+4x-3x-12\)
\(=x\left(x+4\right)-3\left(x+4\right)\)
\(=\left(x+4\right)\left(x-3\right)\)
c) Ta có: \(6x^2-11x-16\)
\(=6\left(x^2-\frac{11}{6}x-\frac{16}{6}\right)\)
\(=6\left(x^2-2\cdot x\cdot\frac{11}{12}+\frac{121}{144}-\frac{505}{144}\right)\)
\(=6\left[\left(x-\frac{11}{12}\right)^2-\frac{505}{144}\right]\)
\(=6\left(x-\frac{11+\sqrt{505}}{12}\right)\left(x-\frac{11-\sqrt{505}}{12}\right)\)
d) Ta có: \(4x^2-8x-5\)
\(=4x^2-10x+2x-5\)
\(=2x\left(2x-5\right)+\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+1\right)\)
e) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x^2-3x-5x+15\right)\)
\(=\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
g) Ta có: \(x^3+9x^2+23x+15\)
\(=x^3+x^2+8x^2+8x+15x+15\)
\(=x^2\left(x+1\right)+8x\left(x+1\right)+15\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+8x+15\right)\)
\(=\left(x+1\right)\left(x^2+3x+5x+15\right)\)
\(=\left(x+1\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
h) Ta có: \(2x^4-x^3-9x^2+13x\)
\(=x\left(2x^3-x^2-9x+13\right)\)
i) Ta có: \(x^4+2x^3-16x^2-2x+15\)
\(=x^4-3x^3+5x^3-15x^2-x^2+3x-5x+15\)
\(=x^3\left(x-3\right)+5x^2\left(x-3\right)-x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3+5x^2-x-5\right)\)
\(=\left(x-3\right)\left[x^2\left(x+5\right)-\left(x+5\right)\right]\)
\(=\left(x-3\right)\left(x+5\right)\left(x^2-1\right)\)
\(=\left(x-3\right)\left(x+5\right)\left(x-1\right)\left(x+1\right)\)
a) \(5xy^3+30x^2z^2-6x^3yz-25y^2z\)
\(=\left(5xy^3-25y^2z\right)+\left(30x^2z^2-6x^3yz\right)\)
\(=5y^2\left(xy-5z\right)+6x^2z\left(5z-xy\right)\)
\(=5y^2\left(xy-5z\right)-6x^2z\left(xy-5z\right)\)
\(=\left(xy-5z\right)\left(5y^2-6x^2z\right)\)
P/s:Bài này chỉ có nước mỏ hạng tử để ghép -_-
xí câu dễ nhất (còn lại làm sau)
c): Đặt \(t=x^2-x\). Ta có:
\(\left(t+1\right)\left(t+2\right)-12=t^2+3t-10\)
\(=t^2+5t-2t-10\)
\(=t\left(t+5\right)-2\left(t+5\right)=\left(t-2\right)\left(t+5\right)\)
\(=\left(x^2-x-2\right)\left(x^2-x+5\right)\)