x^3-x^2+xy^2-4xy+y^3+5
tại x-y=4
Giúp với mn >:((
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\(4xy\left(x^2+y^2\right)-6\left(x^3+y^3+x^2y+xy^2\right)+9\left(x^2+y^2\right)\)
\(=4xy\left(x^2+y^2\right)-6\left[x\left(x^2+y^2\right)+y\left(x^2+y^2\right)\right]+9\left(x^2+y^2\right)\)
\(=4xy\left(x^2+y^2\right)-6\left(x^2+y^2\right)\left(x+y\right)+9\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)\left(4xy-6x-6y+9\right)\)
\(=\left(x^2+y^2\right)\left[2x\left(2y-3\right)-3\left(2y-3\right)\right]\)
\(=\left(x^2+y^2\right)\left(2y-3\right)\left(2x-3\right)\)
\(\dfrac{x}{2}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{20}=\dfrac{z}{12}\)
Áp dụng t/c của dãy số bằng nhau, ta có: \(\dfrac{x-y+z}{10-20+12}=\dfrac{4}{2}=2\)
\(\dfrac{x}{10}=2\Rightarrow x=20\)
\(\dfrac{y}{20}=2\Rightarrow y=40\)
\(\dfrac{z}{12}=2\Rightarrow z=24\)
x/10=y/20=z/12
x-y+z/=10-20+12=4/2=2
x=2.10=20
y=2.20=40
z=2.12=24
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)