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\(A=3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)
\(=3x^2+3y^2-2\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=3x^2+3y^2-2.1\left(x^2-xy+y^2\right)\)
\(=3x^2+3y^2-2x^2+2xy-2y^2\)
\(=x^2+2xy+y^2=\left(x+y\right)^2=1^2=1\)
\(B=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2.1\)
\(=x^3+y^3+3xy\left(x+y\right)^2-6x^2y^2+6x^2y^2\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=x^2-xy+y^2+3xy\)
\(=x^2+2xy+y^2=\left(x+y\right)^2=1^2=1\)
\(x+y=1\)
\(\Leftrightarrow\)\(\left(x+y\right)^2=1\)
\(\Leftrightarrow\)\(x^2+y^2=1-2xy\)
\(x+y=1\)
\(\Leftrightarrow\)\(\left(x+y\right)^3=1\)
\(\Leftrightarrow\)\(x^3+y^3=1-3xy\)
\(H=1-3xy+3xy\left(1-2xy\right)+6x^2y^2\left(xy+y\right)\)
\(=1-6x^2y^2+6x^2y^2\left(xy+y\right)\)
\(=1-6x^2y^2\left(1-xy-y\right)\)
\(=1-6x^2y^2\left(x+y-xy-y\right)\)
\(=1-6x^2y^2\left(x-xy\right)\)
\(=1-6x^3y^2\left(1-y\right)\)
\(=1-6x^3y^2\left(x+y-y\right)\)
\(=1-6x^4y^2\)
mới ra đc đến đây
\(N=x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)
\(N=x^3+y^3+6x^2y^2+3xy\left[\left(x+y\right)^2-2xy\right]\)
\(N=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2+3xy-6x^2y^2\)
\(N=x^2-xy+y^2+3xy\)
\(N=\left(x+y\right)^2\)
\(N=1\)
\(x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2\left(x+y\right)+3xy\left[\left(x+y\right)^2-2xy\right]\)
\(=x^2-xy+y^2+6x^2y^2+3xy-6x^2y^2\)( Do \(x+y=1\))
\(=\left(x+y\right)^2-2xy-xy+3xy+6x^2y^2-6x^2y^3\)
\(=\left(x+y\right)^2=1^2=1\)
`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`
`= 2xy`.
Thay `x = 2/3; y = -3/4` vào BT:
`2 . 2/3 . -3/4 = -1.`
`b, x(x-2y) - y(y^2-2x)`
`= x^2 - 2xy - y^3 + 2xy`
`= x^2 - y^3`
Thay `x = 5; y =3` vào BT:
`= 5^2 - 3^3 = 25 - 27 = -2`
a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)
\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)
\(=2xy\)
Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:
\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)
b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)
\(=x^2-2xy-y^3+2xy\)
\(=x^2-y^3\)
Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)
Ta có : \(x^2-x=y^2-y\)
\(\Leftrightarrow x^2-x-y^2+y=0\)
\(\Leftrightarrow x^2-y^2-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)
Do \(x;y\) khác nhau
\(\Rightarrow x-y\ne0\)
\(\Rightarrow x+y-1=0\)
\(\Rightarrow x+y=1\)
Lại có : \(B=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x^2+y^2\right)+6x^2y^2\)
\(=x^2-xy+y^2+3xy\left(x^2+y^2+2xy\right)\)
\(=x^2-xy+y^2+3xy\left(x+y\right)^2\)
\(=x^2-xy+y^2+3xy\)
\(=x^2+2xy+y^2\)
\(=\left(x+y\right)^2\)
\(=1\)
Vậy \(B=1\)
Sửa đề: x+y=1
\(A=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\)
\(=1-3xy+3xy\left[1-2xy\right]+6x^2y^2\)
=1
a) Ta có:
\(x+y=1\)
\(\Rightarrow\left(x+y\right)^3=1\)
\(\Rightarrow x^3+y^3+3xy\left(x+y\right)=1\)
\(\Rightarrow x^3+y^3+3xy=1\)
\(\Rightarrow P=1\)
b) \(Q=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
\(Q=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\left(x+y\right)\)
\(Q=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\left(x+y\right)\)
Thay x + y = 1 vào Q
\(Q=1-3xy+3xy\left(1-2xy\right)+6x^2y^2\)
\(Q=1-3xy+3xy-6x^2y^2+6x^2y^2\)
\(Q=1\)