5x\(^2\) + 5xy - x - y, phân tích thành nhân tử
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2+5x-6\\ =x^2-x+6x-6\\ =x\left(x-1\right)+6\left(x-1\right)\\ =\left(x+6\right)\left(x-1\right)\\ ---\\ 5x^2+5xy-x-y\\ =x\left(5x-1\right)+y\left(5x-1\right)\\ =\left(5x-1\right)\left(x+y\right)\\ ----\\ 7x-6x^2-2\\ =-6x^2+3x+4x-2\\ =-3x\left(2x-1\right)+2\left(2x-1\right)\\ =\left(2-3x\right)\left(2x-1\right)\)
\(x^2+5x-6=\left(x-1\right)\left(x+6\right)\\ 5x^2+5xy-x-y=\left(5x-1\right)\left(x+y\right)\\ 7x-6x^2-2=\left(3x-2\right)\left(2x-1\right)\)
`@` `\text {Ans}`
`\downarrow`
`x^2 + xy - 2x - 2y`
`= (x^2 - 2x) + (xy - 2y)`
`= x(x - 2) + y(x - 2)`
`= (x + y)(x - 2)`
____
`x^2 - xy - 6x + 6y`
`= (x^2 - 6x) - (xy - 6y)`
`= x(x - 6) - y(x - 6)`
`= (x - y)(x - 6)`
____
`5xy^2 - 5x + y^2 - 1`
`= (5xy^2 + y^2) - (5x + 1)`
`= y^2(5x + 1) - (5x + 1)`
`= (y^2 - 1)(5x + 1)`
`= (y - 1)(y + 1)(5x + 1)`
a: =(x^2+xy)-(2x+2y)
=x(x+y)-2(x+y)
=(x+y)(x-2)
b: =(x^2-xy)-(6x-6y)
=x(x-y)-6(x-y)
=(x-y)(x-6)
c: =5xy^2+y^2-5x-1
=y^2(5x+1)-(5x+1)
=(5x+1)(y^2-1)
=(5x+1)(y+1)(y-1)
\(x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x+6\right)\left(x-1\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=3x-6x^2+4x-2=3x\left(1-2x\right)-2\left(1-2x\right)=\left(3x-2\right)\left(1-2x\right)\)
\(a,x^2+5x-6=x^2-x+6x-6\)
\(=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
\(b,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
\(c,7x-6x^2-2=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2x-1\right)\left(2-3x\right)\)
5x2 - 5xy + 4y - 4x
= 5x ( x - y ) - 4 ( x - y )
= ( 5x - 4 ) ( x - y )
( x + y )3 + ( x - y )3
= 2x3 + 6xy2
= 2x ( x2 + 3y2 )
5 x^2 - 5xy + 4y - 4x
= 5x ( x - y ) - 4 ( x - y )
= ( x - y ) ( 5x - 4 )
( x + y )^3 + ( x - y )^3
= \(x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)
= \(2x^3+6xy^2\)
=\(2x\left(x^2+3y^2\right)\)
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a) \(5x^2+5xy-x-y\)
\(=5x.\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
b) \(5x^2-10y+5y^2-20z^2\)
\(=5.\left(x^2-2y+y^2-4z^2\right)\)
Đề sai ở đâu đó.
c) \(4x^2-y^2+4x+1\)
\(=\left(4x+4x^2+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+y+1\right)\left(2x-y+1\right)\)
a, 5x^2 +5xy - x - y
= 5x ( x+ y ) - (x + y)
= ( 5x - 1)(x + y)
b, 2x^2 + 3x - 5
= 2x^2 - 2x + 5x - 5
= 2x( x - 1) + 5( x - 1)
= ( 2x + 5 )(x- 1 )
c; 16x - 5x^2 - 3
c, = - ( 5x^2 - 16x + 3 )
= - ( 5x^2 - x - 15x + 3 )
= - [ x(5x - 1 ) - 3 (5x - 1) ]
= - ( x- 3)(5x - 1 )
`5x^2 + 5xy - x - y`
`= 5x(x+y)-(x+y)`
`= (5x-1)(x+y)`