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A=3.(5-xy)
ta có: \(\left(x+y\right)^2=9\Leftrightarrow x^2+2xy+y^2=9\Leftrightarrow5+2xy=9\Leftrightarrow xy=2\)
=> A=3(5-2)=9
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1, Ta có \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)
\(\Rightarrow\) \(a^3+b^3+c^3=0.\left(a^2+b^2+c^2-ab-bc-ac\right)+3.\left(-2\right)=-6\)
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ta có:(x+y)^2=1 (vì x+y=1)
=> x^2+2xy+y^2=1
=>x^2+y^2=1-2xy
=>1-2xy=85 (vì x^2+y^2=85)
=>2xy=-84
=>xy=-42 (1)
Ta có: x^3+y^3=(x^3+3x^2y+3xy^2+y^3)-3x^2y+3xy^2 (thêm bớt 3x^2y và 3xy^2)
=(x+y)^3-3xy(x+y)
=1-3xy (2)
Từ (1) va (2) =>x^3+y^3=1-(-126)=127
Vậy x^3+y^3=127
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Xét hiệu
\(\left(x^3+y^3+z^3\right)-\left(x+y+z\right)\\ =\left(x^3-x\right)+\left(y^3-y\right)+\left(z^3-z\right)\\ =\left(x-1\right)x\left(x+1\right)+\left(y-1\right)y\left(y+1\right)+\left(z-1\right)z\left(z+1\right)⋮6\)
Mà\(x+y+z⋮6\)
\(\Rightarrow x^3+y^3+z^3⋮6\)
\(x^3+y^3=108\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)=108\)
\(\Rightarrow x^2-xy+y^2=18\left(1\right)\)
\(x+y=6\Rightarrow\left(x+y\right)^2=36\Leftrightarrow x^2+2xy+y^2=36\left(2\right)\)
Lấy (2)-(1) ta được 3xy=18=>xy=6
mình ko hỉu đề