Cho x, y phân biệt thỏa mãn 1/x^2+1 + 1/ y^2+1 = z/xy+1 Tính H = 1/ x^2+1 + 1/y^2+1 +2/xy+1 24/07/2021 Bởi Allison Cho x, y phân biệt thỏa mãn 1/x^2+1 + 1/ y^2+1 = z/xy+1 Tính H = 1/ x^2+1 + 1/y^2+1 +2/xy+1
Đáp án: $H=2$ Giải thích các bước giải: $\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{xy+1}$ $\rightarrow \dfrac{1}{x^2+1}-\dfrac{1}{xy+1}+\dfrac{1}{y^2+1}-\dfrac{1}{xy+1}=0$ $\rightarrow \dfrac{xy-x^2}{(x^2+1)(xy+1)}+\dfrac{xy-y^2}{(y^2+1)(xy+1)}=0$ $\rightarrow \dfrac{x(y-x)}{(x^2+1)(xy+1)}+\dfrac{y(x-y)}{(y^2+1)(xy+1)}=0$ $\rightarrow \dfrac{x}{x^2+1}-\dfrac{y}{y^2+1}=0$ $\rightarrow \dfrac{x}{x^2+1}=\dfrac{y}{y^2+1}$ $\rightarrow x(y^2+1)=y(x^2+1)$ $\rightarrow x^2y-xy^2+y-x=0$ $\rightarrow xy(x-y)-(x-y)=0$ $\rightarrow(x-y)(xy-1)=0$ $\rightarrow xy-1=0$ $\rightarrow xy=1$ $\rightarrow \dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{xy+1}=1$ $\rightarrow H=2$ Bình luận
Đáp án: $H=2$
Giải thích các bước giải:
$\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{xy+1}$
$\rightarrow \dfrac{1}{x^2+1}-\dfrac{1}{xy+1}+\dfrac{1}{y^2+1}-\dfrac{1}{xy+1}=0$
$\rightarrow \dfrac{xy-x^2}{(x^2+1)(xy+1)}+\dfrac{xy-y^2}{(y^2+1)(xy+1)}=0$
$\rightarrow \dfrac{x(y-x)}{(x^2+1)(xy+1)}+\dfrac{y(x-y)}{(y^2+1)(xy+1)}=0$
$\rightarrow \dfrac{x}{x^2+1}-\dfrac{y}{y^2+1}=0$
$\rightarrow \dfrac{x}{x^2+1}=\dfrac{y}{y^2+1}$
$\rightarrow x(y^2+1)=y(x^2+1)$
$\rightarrow x^2y-xy^2+y-x=0$
$\rightarrow xy(x-y)-(x-y)=0$
$\rightarrow(x-y)(xy-1)=0$
$\rightarrow xy-1=0$
$\rightarrow xy=1$
$\rightarrow \dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}=\dfrac{2}{xy+1}=1$
$\rightarrow H=2$