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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
Bài 2:
a: \(3\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)\)
\(=3\left(x^3-1\right)+x^3-3x^2+3x-1-4x\left(x^2-1\right)\)
\(=3x^3-3+x^3-3x^2+3x-1-4x^3+4x\)
\(=-3x^2+7x-4\)
\(=-3\cdot\left(-1\right)^2+7\cdot\left(-1\right)-4\)
=-3-4-7=-14
b: \(=27x^3y^3-8-3xy\left(9x^2y^2+6xy+1\right)\)
\(=27x^3y^3-8-27x^3y^3-18x^2y^2-3xy\)
\(=-18x^2y^2-3xy-8\)
\(=-18\cdot\left[\left(-2010\right)\cdot\left(-\dfrac{1}{2010}\right)\right]^2-3\cdot\left(-2010\right)\cdot\dfrac{-1}{2010}-8\)
\(=-18-3-8=-29\)
A = (3x + y)^2 - 3y . ( 2x - 1/3y )
=2y2+9x2
B = ( x - 2 )^2 + ( x + 2 )^2 - 2. ( x - 2 ) ( x + 2)
=24
C = ( x - y ) ( x^2 + xy + y^2 ) + 2y^3
=y3+x3
D = ( x -5 ) ( x+ 5 ) -(x - 8 ) (x + 4)
=4x+7
E = (3x + 1 )^2 - 2 . ( 9x^2 - 1 ) + ( 3x - 1 ) ^2
=4
F = ( x - 3 ) ( x + 3 ) - ( x - 3 )^2
=6x-18
a) \(\dfrac{x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^3-1}\)
\(=\dfrac{\left(x^8+x^7+x^6\right)+\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)}{x^3-1}\)
\(=\dfrac{x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)}{x^3-1}\)
\(=\dfrac{\left(x^2+x+1\right)\left(x^6+x^3+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^6+x^3+1}{x-1}\)
b) \(\dfrac{x^5+x+1}{x^3+x^2+x}\)
\(=\dfrac{x^5+x^4+x^3+x^2-x^4-x^3-x^2+x+1}{x^3+x^2+x}\)
\(=\dfrac{\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)}{x^3+x^2+x}\)
\(=\dfrac{x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}{x^3+x^2+x}\)
\(=\dfrac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}\)
\(=\dfrac{x^3-x^2+1}{x}\)