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a) M = (x + 3y)² - (x - 3y)²
= [(x + 3y) - (x - 3y)][(x + 3y) + (x - 3y)]
= (x + 3y - x + 3y)(x + 3y + x - 3y)
= 6y.2x
= 12xy
b) Q = (x - y)² - 4(x - y)(x + 2y) + 4(x + 2y)²
= [(x - y) - 2(x + 2y)]²
= (x - y - 2x - 4y)²
= (-x - 5y)²
a) M = (x + 3y)² - (x - 3y)²
= [(x + 3y) - (x - 3y)][(x + 3y) + (x - 3y)]
= (x + 3y - x + 3y)(x + 3y + x - 3y)
= 6y.2x
= 12xy
b: Q=(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2
=(x-y-2x-4y)^2
=(-x-5y)^2
=x^2+10xy+25y^2
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
a)\(A=\left(\frac{x+y}{x-2y}+\frac{3y}{2y-x}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(\frac{x+y-3y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(\frac{x-2y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(1-3xy\right).\frac{-x-1}{1-3xy}+\frac{x^2}{x+1}\)
\(=-\left(x+1\right)+\frac{x^2}{x+1}\)`
\(=\frac{-\left(x+1\right)^2+x^2}{x+1}\)
\(=\frac{-x^2-2x-1+x^2}{x+1}\)
\(=\frac{-2x-1}{x+1}\)(1)
b) Thay \(x=-3,y=2014\)vào (1) ta được:
\(A=\frac{-2.\left(-3\right)-1}{-3+1}=\frac{-5}{2}\)
Vậy \(A=\frac{-5}{2}\)với x=-3 và y=2014
1. \(\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2\)
\(=x^2-y^2+x^2+2xy+y^2\)
\(=2x^2+2xy\)
2. \(\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(B=\left(x+3y\right)\left(x-3y\right)-y\left(x+9y\right)\)
\(=x^2-9y^2-xy-9y^2\)
\(=x^2-xy\)
\(C=\left(3x-9\right)\left(x^2+3x+9\right)-3x\left(x^2-2\right)\)
\(=3x^3-81-3x^3+6x\)
=6x-81