Cho:
\(A=x^3+x^2y-5^2+10xy+7y^3+8.\)
\(B=y^3-3x^2y+6y^2+2xy^2-19xy-8x^3+2.\)
Tính:A+B
A-B
B-A
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\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{\left(x-y\right)+\left(x+y\right)}{3+13}=\frac{2x}{16}=\frac{x}{8}=\frac{25x}{200}=\frac{xy}{200}\)
\(\Rightarrow y=25\)
\(\frac{x-25}{3}=\frac{x+25}{13}\)
\(\Leftrightarrow13\left(x-25\right)=3\left(x+25\right)\)
\(\Leftrightarrow13x-325=3x+75\)
\(\Leftrightarrow13x-3x=75+325\)
\(\Leftrightarrow10x=400\)
\(\Rightarrow x=40\)
Vậy \(x=40;y=25\)
Ta có :
\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{b}{a+c}+\frac{a+c}{a+c}\right)+\left(\frac{c}{a+b}+\frac{a+b}{a+b}\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=2009.\frac{1}{7}=287\)
\(\Rightarrow S=287-3=284\)
\(A+B=x^3+x^2y-5^2+10xy+7y^3+8+y^3-3x^2y+6y^2+2xy^2-19xy-8x^3+2.\)
\(=x^3-8x^3+x^2y-3x^2y+10xy-19xy+7y^3+y^3+6y^2+2xy^2-25+8+2.\)
\(=-7x^3-2x^2y-9xy+8y^3+2y^2\left(3+x\right)-15.\)