Question

1/Cho x+y=9; xy=18. Tính giá trị A=x³-y³

2/Cho \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\); \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\). Tính \(M=\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}\)

Answer 1

1, Ta có: \(x+y=9\Rightarrow\left(x+y\right)^2=81\)

\(\Rightarrow x^2+2xy+y^2=81\)

\(\Rightarrow x^2+y^2=45\)

\(\Rightarrow x^2+y^2-2xy=9\)

\(\Rightarrow\left(x-y\right)^2=9\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-3\end{matrix}\right.\)

\(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(\Rightarrow\left[{}\begin{matrix}A=3.63=189\\A=-3.63=-189\end{matrix}\right.\)

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