Toán Cho x+ $\frac{1}{y}$ =3 ; y+ $\frac{1}{x}$ =5. Tính $\frac{x^3y^3}{x^6y^6+1}$ 09/09/2021 By Adalynn Cho x+ $\frac{1}{y}$ =3 ; y+ $\frac{1}{x}$ =5. Tính $\frac{x^3y^3}{x^6y^6+1}$
Đáp án: \(\dfrac{x^3y^3}{x^6y^6+1}=\dfrac1{2158}\) Giải thích các bước giải: \(\begin{cases}x+\dfrac1y=3\\ y+\dfrac1x=5\end{cases}\) \(\Rightarrow \left(x+\dfrac1y\right)\left(y+\dfrac1x\right)=3\cdot 5=15\\ \Rightarrow xy+1+\dfrac 1{xy}+1=15\\ \Rightarrow xy+\dfrac 1{xy}=13\\ \Rightarrow \left(xy+\dfrac1{xy}\right)^3=13^3\\ \Rightarrow x^3y^3+3x^2y^2\cdot \dfrac1{xy}+3xy\cdot \dfrac1{x^2y^2}+\dfrac1{x^3y^3}=2197\\ \Rightarrow x^3y^3+\dfrac1{x^3y^3}+3\left(xy+\dfrac1{xy}\right)=2197\\ \Rightarrow x^3y^3+\dfrac1{x^3y^3}=2197-3\cdot 13=2158\\ \Rightarrow \dfrac{x^6y^6+1}{x^3y^3}=2158\\ \Rightarrow \dfrac{x^3y^3}{x^6y^6+1}=\dfrac1{2158}\) Vậy \(\dfrac{x^3y^3}{x^6y^6+1}=\dfrac1{2158}\) Trả lời
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Đáp án:
\(\dfrac{x^3y^3}{x^6y^6+1}=\dfrac1{2158}\)
Giải thích các bước giải:
\(\begin{cases}x+\dfrac1y=3\\ y+\dfrac1x=5\end{cases}\)
\(\Rightarrow \left(x+\dfrac1y\right)\left(y+\dfrac1x\right)=3\cdot 5=15\\ \Rightarrow xy+1+\dfrac 1{xy}+1=15\\ \Rightarrow xy+\dfrac 1{xy}=13\\ \Rightarrow \left(xy+\dfrac1{xy}\right)^3=13^3\\ \Rightarrow x^3y^3+3x^2y^2\cdot \dfrac1{xy}+3xy\cdot \dfrac1{x^2y^2}+\dfrac1{x^3y^3}=2197\\ \Rightarrow x^3y^3+\dfrac1{x^3y^3}+3\left(xy+\dfrac1{xy}\right)=2197\\ \Rightarrow x^3y^3+\dfrac1{x^3y^3}=2197-3\cdot 13=2158\\ \Rightarrow \dfrac{x^6y^6+1}{x^3y^3}=2158\\ \Rightarrow \dfrac{x^3y^3}{x^6y^6+1}=\dfrac1{2158}\)
Vậy \(\dfrac{x^3y^3}{x^6y^6+1}=\dfrac1{2158}\)