x^3-x^2+xy^2-4xy+y^3+5
tại x-y=4
Giúp với mn >:((
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\(x=\dfrac{1}{y}\Rightarrow\dfrac{1}{y}-y=4\\ \Rightarrow y^2+4y-1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\\y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\end{matrix}\right.\)
Với \(x=2-\sqrt{5};y=-2-\sqrt{5}\)
\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^2=322\)
Với \(x=2+\sqrt{5};y=-2+\sqrt{5}\)
\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^4=322\)
A=x^2+y^2
=(x-y)^2+2xy
=4^2+2=18
B=(x-y)^3+3xy(x-y)
=4^3+3*1*4
=64+12=76
C=(x^2+y^2)^2-2x^2y^2
=18^2-2
=322
\(4xy\left(x^2+y^2\right)-6\left(x^3+y^3+x^2y+xy^2\right)+9\left(x^2+y^2\right)\)
\(=4xy\left(x^2+y^2\right)-6\left[x\left(x^2+y^2\right)+y\left(x^2+y^2\right)\right]+9\left(x^2+y^2\right)\)
\(=4xy\left(x^2+y^2\right)-6\left(x^2+y^2\right)\left(x+y\right)+9\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)\left(4xy-6x-6y+9\right)\)
\(=\left(x^2+y^2\right)\left[2x\left(2y-3\right)-3\left(2y-3\right)\right]\)
\(=\left(x^2+y^2\right)\left(2y-3\right)\left(2x-3\right)\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
\(\dfrac{x}{2}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{20}=\dfrac{z}{12}\)
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
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)