x3-3xy-10y2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(x^2\) + 4\(x\) - y2 + 4
= (\(x^2\) + 4\(x\) + 4) - y2
= (\(x\) + 2)2 - y2
= (\(x\) + 2 - y)(\(x\) + 2 + y)
b, 2\(x^2\) - 18
= 2.(\(x^2\) -9)
= 2.(\(x\) -3).(\(x\) + 3)
9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)
a) (a + b)2 – m2 + a + b – m = (a + b + m)(a + b – m) + (a + b – m)
= (a + b – m)(a + b + m + 1)
b) x3 + 6x2 + 12x – 8 = (x – 2)3
Cách khác: x3 + 6x2 + 12x – 8 = (x3 – 8) - 6x2 + 12x
= (x – 2)(x2 + 2x + 2) + 6x(x – 2) = (x – 2)( x2 + 2x + 2 – 6x) = (x – 2)3
c) x2 – 7xy + 10y2 = x2 – 2xy – 5xy + 10y2 = x(x – 2y) – 5y(x – 2y)
= (x – 2y)(x – 5y)
d) x4 + 2x3 - 4x – 4 = (x4 – 4) + (2x3 – 4x) = (x2 – 2)(x2 + 2) + 2x(x2 – 2)
= (x2 – 2)(x2 + 2 + 2x) = (x - √2)( x + √2)( x2 + 2 + 2x)
học tốt
a,(a+b+m)(a+b-m)+(a+b-m)
=(a+b-m)(a+b+m+1)
...........mấy câu kia tương tự:>
hc tốt
a) \(A=x^3+y^3+3xy\)
\(=x^3+y^3+3xy\left(x+y\right)\) (do \(x+y=1\))
\(=x^3+3x^2y+3xy^2+y^3\)
\(=\left(x+y\right)^3\) \(=1\)
b) \(B=x^3-y^3-3xy\)
\(=x^3-y^3-3xy\left(x-y\right)\) (do \(x-y=1\))
\(=x^3-3x^2y+3xy^2-y^3\)
\(=\left(x-y\right)^3\) \(=1\)
Q = x - y 3 + y + x 3 + y - x 3 – 3xy(x + y)
= x 3 – 3 x 2 y + 3x y 2 – y 3 + y 3 + 3 y 2 .x + 3y x 2 + x 3 + y 3 – 3 y 2 .x +3y x 2 – x 3 – 3 x 2 y – 3x y 2
= x 3 – 3 x 2 y + 3x y 2 – y 3 + y 3 + 3.x y 2 + 3 x 2 .y + x 3 + y 3 – 3x. y 2 + 3 x 2 .y – x 3 – 3 x 2 y – 3x y 2
= ( x 3 + x 3 – x 3 )+ ( - 3 x 2 y + 3 x 2 y+ 3 x 2 y – 3 x 2 y)+ (3x y 2 + 3x y 2 - 3x y 2 - 3x y 2 ) + (- y 3 + y 3 + y 3 )
= x 3 + 0 x 2 y + 0.x y 2 + y 3
= x 3 + y 3
`#3107.101107`
`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`
Ta có:
`x - y - 1 = 0`
`=> x - y = 1`
`D = x^3 - y^3 - 3xy`
`= (x - y)(x^2 + xy + y^2) - 3xy`
`= 1 * (x^2 + xy + y^2) - 3xy`
`= x^2+ xy + y^2 - 3xy`
`= x^2 - 2xy + y^2`
`= x^2 - 2*x*y + y^2`
`= (x - y)^2`
`= 1^2 = 1`
Vậy, với `x - y = 1` thì `D = 1`
________
`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`
`x + y = 5`
`=> (x + y)^2 = 25`
`=> x^2 + 2xy + y^2 = 25`
`=> 2xy = 25 - (x^2 + y^2)`
`=> 2xy = 25 - 17`
`=> 2xy = 8`
`=> xy = 4`
Ta có:
`E = x^3 + y^3`
`= (x + y)(x^2 - xy + y^2)`
`= 5 * [ (x^2 + y^2) - xy]`
`= 5 * (17 - 4)`
`= 5 * 13`
`= 65`
Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`
________
`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`
Ta có:
`x - y = 4`
`=> (x - y)^2 = 16`
`=> x^2 - 2xy + y^2 = 16`
`=> (x^2 + y^2) - 2xy = 16`
`=> 2xy = (x^2 + y^2) - 16`
`=> 2xy = 26 - 16`
`=> 2xy = 10`
`=> xy = 5`
Ta có:
`F = x^3 - y^3`
`= (x - y)(x^2 + xy + y^2)`
`= 4 * [ (x^2 + y^2) + xy]`
`= 4 * (26 + 5)`
`= 4*31`
`= 124`
Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`
\(a,x+y=1\Leftrightarrow\left(x+y\right)^3=1\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\\ \Leftrightarrow x^3+y^3+3xy\cdot1=1\Leftrightarrow x^3+y^3+3xy=1\)
\(b,x^3-y^3-3xy\\ =x^3-3x^2y+3xy^2-y^3-3xy+3x^2y-3xy^2\\ =\left(x-y\right)^3-3xy\left(x-y-1\right)\\ =1^3-3xy\left(1-1\right)=1-0=1\)
\(c,x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\\ =x^2-xy+y^2+3xy-6x^2y^2+6x^2y^2\\ =x^2+2xy+y^2=\left(x+y\right)^2=1\)
a: =(x-1)^3
b: =(-2x+1)^3
c: =x^3-3x^2y+3xy^2-y^3
=(x-y)^3
\(x^2-3xy-10y^2=x^2-5xy+2xy-10y^2=x\left(x-5y\right)+2y\left(x-5y\right)=\left(x-5y\right)\left(x+2y\right)\)