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a) \(P=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)-100\)
\(P=3\left(x+y\right)^2-2.5-100\)
\(P=3.5^2-110\)
\(P=-35\)
b) \(Q=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3.5+10\)
\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+25\)
\(Q=5^3-2.5^2+25\)
\(Q=100\)
\(x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3\left(x+y\right)+10=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10=5^3-2.5^2+3.5+10=100\)
\(a,A=x^2+y^2\\=x^2-2xy+y^2+2xy\\=(x-y)^2+2xy\\=2^2+2\cdot1\\=4+2\\=6\)
\(b,x+y=1\\\Leftrightarrow (x+y)^3=1^3\\\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\\\Leftrightarrow x^3+3xy(x+y)+y^3=1\\\Leftrightarrow x^3+3xy\cdot1+y^3=1\\\Rightarrow A=1\)
a) Ta có:
\(x-y=2\)
\(\Rightarrow\left(x-y\right)^2=2^2\)
\(\Rightarrow x^2-2xy+y^2=4\)
Mà: \(xy=1\)
\(\Rightarrow\left(x^2+y^2\right)-2\cdot1=4\)
\(\Rightarrow x^2+y^2=4+2\)
\(\Rightarrow x^2+y^2=6\)
b) Ta có:
\(x+y=1\)
\(\Rightarrow\left(x+y\right)^3=1^3\)
\(\Rightarrow x^3+3x^2y+3xy+y^3=1\)
\(\Rightarrow x^3+3xy\left(x+y\right)+y^3=1\)
Mà: x + y = 1
\(\Rightarrow x^3+3xy\cdot1+y^3=1\)
\(\Rightarrow x^3+3xy+y^3=1\)
`#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) \(A=x^3+y^3+3xy\)
\(=x^3+y^3+3xy\left(x+y\right)\) (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 \(x-y=1\))
\(=x^3-3x^2y+3xy^2-y^3\)
\(=\left(x-y\right)^3\) \(=1\)