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`#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.`
10: \(x\left(x-y\right)+x^2-y^2\)
\(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+x+y\right)\)
\(=\left(x-y\right)\left(2x+y\right)\)
11: \(x^2-y^2+10x-10y\)
\(=\left(x^2-y^2\right)+\left(10x-10y\right)\)
\(=\left(x-y\right)\left(x+y\right)+10\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+10\right)\)
12: \(x^2-y^2+20x+20y\)
\(=\left(x^2-y^2\right)+\left(20x+20y\right)\)
\(=\left(x-y\right)\left(x+y\right)+20\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+20\right)\)
13: \(4x^2-9y^2-4x-6y\)
\(=\left(4x^2-9y^2\right)-\left(4x+6y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)-2\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-2\right)\)
14: \(x^3-y^3+7x^2-7y^2\)
\(=\left(x^3-y^3\right)+\left(7x^2-7y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\cdot\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+7x+7y\right)\)
15: \(x^3+4x-\left(y^3+4y\right)\)
\(=x^3-y^3+4x-4y\)
\(=\left(x^3-y^3\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+4\right)\)
16: \(x^3+y^3+2x+2y\)
\(=\left(x^3+y^3\right)+\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+2\right)\)
17: \(x^3-y^3-2x^2y+2xy^2\)
\(=\left(x^3-y^3\right)-\left(2x^2y-2xy^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)-2xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2-2xy\right)\)
\(=\left(x-y\right)\left(x^2-xy+y^2\right)\)
18: \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x^2-4x+4\right)-y^2\right]\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
\(b,\left(x+2\right)^2-25\)
\(=\left(x+2\right)^2-5^2\)
\(=\left(x-3\right)\left(x+7\right)\)
\(c,36\left(x-y\right)^2\)
\(=36\left(x^2-2xy+y^2\right)\)
\(=36x^2-72xy+36y^2\)
\(d,x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
\(=x^2+2.x.\dfrac{1}{4}+\dfrac{1}{4}^2\)
\(=\left(x+\dfrac{1}{4}\right)^2\)
\(e,2x^4y^3-3x^2y^4+5x^3y^4\)
\(=x^2y^3\left(2x^2-3y+5xy\right)\)
Các câu còn lại làm tương tự, chú ý sd HĐT
\(a,x+y=1\Leftrightarrow\left(x+y\right)^3=1\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\\ \Leftrightarrow x^3+y^3+3xy\cdot1=1\Leftrightarrow x^3+y^3+3xy=1\)
\(b,x^3-y^3-3xy\\ =x^3-3x^2y+3xy^2-y^3-3xy+3x^2y-3xy^2\\ =\left(x-y\right)^3-3xy\left(x-y-1\right)\\ =1^3-3xy\left(1-1\right)=1-0=1\)
\(c,x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\\ =x^2-xy+y^2+3xy-6x^2y^2+6x^2y^2\\ =x^2+2xy+y^2=\left(x+y\right)^2=1\)
Ta có: \(x-y=5\Rightarrow\left(x-y\right)^2=25\)
\(\Rightarrow x^2+y^2-2xy=25\)
\(\Rightarrow15-2xy=25\)
\(\Rightarrow2xy=-10\Rightarrow xy=-5\)
\(M=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=5\left(15-5\right)=5.10=50\)
Sửa đề: \(A=x^3+x^2y-xy^2-y^3+x^2-y^2+2x+2y+3\)
\(A=x^2\left(x+y\right)-y^2\left(x+y\right)+\left(x-y\right)\left(x+y\right)+2x+2y+3\)
\(=-x^2+y^2+\left(-x+y\right)-2+3\)
\(=-\left(x-y\right)\left(x+y\right)-\left(x-y\right)+1\)
\(=\left(x-y\right)\left(-x-y-1\right)+1\)
\(=\left(x-y\right)\left(1-1\right)+1=1\)
\(a,A=x^2+y^2\\=x^2-2xy+y^2+2xy\\=(x-y)^2+2xy\\=2^2+2\cdot1\\=4+2\\=6\)
\(b,x+y=1\\\Leftrightarrow (x+y)^3=1^3\\\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\\\Leftrightarrow x^3+3xy(x+y)+y^3=1\\\Leftrightarrow x^3+3xy\cdot1+y^3=1\\\Rightarrow A=1\)
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
\(\Rightarrow x^3+3xy\cdot1+y^3=1\)
\(\Rightarrow x^3+3xy+y^3=1\)