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a)\(\frac{x^2+xy}{x^2-y^2}=\frac{x\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{x}{x-y}\)
b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{-5x-2}{x^2-4}\)
\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2}{x+2}\)
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x + y = 2 <=> x = 2 - y (1)
x2 + y2 = 10 <=> (x + y).(x - y) = 10 <=> 2(x - y) = 10 <=> x - y = 5 <=> x = 5 + y (2)
Từ (1)(2) suy ra 2 - y = 5 + y <=> 2y = - 3 <=> y = -1,5 => x = 3,5
x3 + y3 = 3,53 + (-1,5)3 = ...
x2 + y2 = 10 <=> ( x + y )2 - 2xy = 10 <=> 4 - 2xy = 10 <=> -2xy = 6 <=> xy = -3
Khi đo : x3 + y3 = ( x + y )3 - 3xy( x + y ) = 8 + 12 = 20
Ta có
\(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
Ta có \(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
\(=\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
\(=\left(x^8-y^8\right)\left(x^8+y^8\right)\)
\(=x^{16}-y^{16}\)