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a: (x^2+x)^2+4x^2+4x-12
=(x^2+x)^2+4(x^2+x)-12
=(x^2+x+6)(x^2+x-2)
=(x^2+x+6)(x+2)(x-1)
b: =(x^2+8x)^2+22(x^2+8x)+105+15
=(x^2+8x)^2+22(x^2+8x)+120
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)
c: =8x^2+12x-2x-3
=(2x+3)(4x-1)
a: =(x^2+x)^2+4(x^2+x)-12
=(x^2+x+6)(x^2+x-2)
=(x^2+x+6)(x+2)(x-1)
b: =(x^2+8x)^2+22(x^2+8x)+120
=(x^2+8x+12)(x^2+8x+10)
=(x+2)(x+6)(x^2+8x+10)
c: =8x^2+12x-2x-3
=(2x+3)(4x-1)
a) \(=\left(6x\right)^2-2.6x.1+1=\left(6x-1\right)^2\)
b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)
c) \(=\left(3x-y\right)^2-25=\left(3x-y-5\right)\left(3x-y+5\right)\)
d) \(=x\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)
a, 7x - 14
= 7(x-2)
b, 2x - 2y + \(x^2\)- xy
= (2x-2y) + (\(x^2\)-xy)
= 2(x-y) + x(x-y)
= (x-y)(2+x)
c, 6x + 12
= 6(x+2)
\(a,=7\left(x-2\right)\\ b,=2\left(x-y\right)+x\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=6\left(x+2\right)\\ d,\text{Sai đề}\)
a) \(=3\left(x^2-10x+25\right)=3\left(x-5\right)^2\)
b) \(=x\left(x+y\right)+8\left(x+y\right)=\left(x+y\right)\left(x+8\right)\)
c) \(=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\)
`a) 8x^2 - 8xy - 4x + 4y`
`= 8x ( x - y ) - 4 ( x - y )`
`= ( x - y ) ( 8x - 4 )`
__________________________
`b) x^3 + 10x^2 + 25x - xy^2`
`=x ( x^2 + 10x + 25 ) - xy^2`
`= x ( x + 5 )^2 - xy^2`
`= x [ ( x + 5 )^2 - y^2 ]`
`= x ( x + 5 - y ) ( x + 5 + y )`
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`c) x^2 + x - 6`
`= x^2 + 3x - 2x - 6`
`= x ( x + 3 ) - 2 ( x + 3 )`
`= ( x + 3 ) ( x - 2 )`
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`d) 2x^2 + 4x - 16`
`= 2x^2 - 4x + 8x - 16`
`= 2x ( x - 2 ) + 8 ( x - 2 )`
`= ( x - 2 ) ( 2x + 8 )`
a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).
b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2
= (a + 4 – b)(a + 4 + b).
tui chỉ làm dc này thui
\(a^3+a+30\)
\(=a^3+3a^2-3a^2-9a+10a+30\)
\(=\left(a+3\right)\left(a^2-3a+10\right)\)
\(x^3+x^2+100\)
\(=x^3+5x^2-4x^2-20x+20x+100\)
\(=\left(x+5\right)\left(x^2-4x+20\right)\)
Bài 1:
\(\left(x^2-y\right)\left(3x+y^2\right)-\left(6x^4y-2xy^4\right):2xy\)
\(=3x\cdot x^2+y^2\cdot x^2-y\cdot3x-y\cdot y^2-6x^4y:2xy+2xy^4:2xy\)
\(=3x^3+x^2y^2-3xy-y^3-3x^3+y^3\)
\(=x^2y^2-3xy\)
Bài 2:
a) \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)
\(=10x^2\left(2x-y\right)-6xy\left(2x-y\right)\)
\(=2x\left(2x-y\right)\left(5x-3y\right)\)
b) \(x^2-2x+1-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-y-1\right)\left(x+y-1\right)\)
c) \(x^2-8x+12\)
\(=x^2-8x+16-4\)
\(=\left(x-4\right)^2-2^2\)
\(=\left(x-6\right)\left(x+2\right)\)
a: \(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
b: \(x^2-1-xy+y\)
\(=\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y+1\right)\)
c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)
\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)
\(=2\cdot\left(x-1\right)\cdot\left(x-2\right)^2\)
a)
ta có \(x^2+8x+7\)
\(=x^2+7x+x+7\)
\(=x\left(x+7\right)+\left(x+7\right)\)
\(=\left(x+7\right)\left(x+1\right)\)
b)
Ta có \(8x^2+30x+7\)
\(=8x^2+2x+28x+7\)
\(=2x\left(4x+1\right)+7\left(4x+1\right)\)
\(\left(4x+1\right)\left(2x+7\right)\)
A)(x+1)(x+7)
B)(x+1/4)(x+7/2)