Rút gọn biểu thức:
A = ( x - y - 1 )3 - ( x - y + 1 )3 + 6 ( x- y )2
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Ta có: \(A=\left(x-y-1\right)^3-\left(x-y+1\right)^3+6\left(x-y\right)^2\)
\(=\left(x-y-1-x+y-1\right)\left[\left(x-y-1\right)^2+\left(x-y-1\right)\left(x-y+1\right)+\left(x-y+1\right)^2\right]+6\left(x-y\right)^2\)
\(=-2\cdot\left[3\left(x-y\right)^2+1\right]+6\left(x-y\right)^2\)
\(=-6\left(x-y\right)^2+6\left(x-y\right)^2-2\)
=-2
a: \(A=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{2}}+y^{\dfrac{1}{3}}\cdot x^{\dfrac{1}{2}}}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=\dfrac{x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}\left(x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}\right)}{x^{\dfrac{1}{6}}+y^{\dfrac{1}{6}}}=x^{\dfrac{1}{3}}\cdot y^{\dfrac{1}{3}}=\left(xy\right)^{\dfrac{1}{3}}\)
b: \(B=\dfrac{x^{3+\sqrt{3}}}{y^2}\cdot\dfrac{x^{-\sqrt{3}-1}}{y^{-2}}=\dfrac{x^{3+\sqrt{3}-\sqrt{3}-1}}{y^{2-2}}=x^2\)
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
a) M = (x² + 3xy - 3x³) + (2y³ - xy + 3x³)
= x² + 3xy - 3x³ + 2y³ - xy + 3x³
= x² + (3xy - xy) + (-3x³ + 3x³) + 2y³
= x² + 2xy + 2y³
Tại x = 5 và y = 4
M = 5² + 2.5.4 + 2.4³
= 25 + 40 + 2.64
= 65 + 128
= 193
b) N = x²(x + y) - y(x² - y²)
= x³ + x²y - x²y + y³
= x³ + (x²y - x²y) + y³
= x³ + y³
Tại x = -6 và y = 8
N = (-6)³ + 8³
= -216 + 512
= 296
c) P = x² + 1/2 x + 1/16
= (x + 1/2)²
Tại x = 3/4 ta có:
P = (3/4 + 1/2)² = (5/4)² = 25/16
\(A,xy\left(2x^2-3\right)-x^2\left(5xy+y\right)+x^2y\\ =2x^3y-3xy-5x^3y-x^2y+x^2y\\ =\left(2x^3y-5x^3y\right)+\left(-x^2y+x^2y\right)-3xy\\ =-3x^3y-3xy\)
\(B,3xyz\left(y-2\right)-5yz\left(1-y\right)-8z\left(y^2-3\right)\\ =3xy^2z-6xyz-5yz+5y^2z-8y^2z+24z\\ =3xy^2z-6xyz+\left(5y^2z-8y^2z\right)-5yz+24z\\ =3xy^2z-6xyz-3y^2z-5yz+24z\)
a) \(27\left(1-x\right)\left(x^2+x+1\right)+81x\left(x-1\right)\)
\(=27\left(1-x^3\right)+81\left(x^2-x\right)\)
\(=27-27x^3+81x^2-81x\)
b) \(y\left[x^2+x\left(x-y\right)+\left(x-y\right)^2\right]+\left(x-y\right)^3\)
\(=y\left[x^2+x^2-xy+x^2-2xy+y^2\right]+x^3-3x^2y+3xy^2-y^3\)
\(=y\left(3x^2-3xy+y^2\right)+x^3-3x^2y+3xy^2-y^3\)
\(=3x^2y-3xy^2+y^3+x^3-3x^2y+3xy^2-y^3=x^3\)