Cho $\frac{x}{a}-\frac{y}{b}-\frac{z}{c}=-2$ và $\frac{a}{x}-\frac{b}{y}-\frac{c}{z}=0$. Tính $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}$
Tính $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}$
2 votes
Started By yellow, 27-12-2012 - 15:44
#1
Posted 27-12-2012 - 15:44
yellow
-
- Pre-Member
-
- 371 posts
Sĩ quan
$\large{\int_{0}^{\infty }xdx<\heartsuit}$
#2
Posted 27-12-2012 - 16:30
Zaraki
-
- Phó Quản lý Toán Cao cấp
-
- 4273 posts
PQT
Cho $\frac{x}{a}-\frac{y}{b}-\frac{z}{c}=-2$ và $\frac{a}{x}-\frac{b}{y}-\frac{c}{z}=0$. Tính $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}$
Ta có $$\begin{aligned} \frac xa - \frac yb - \frac zc =-2 & \Rightarrow \left( \frac xa - \frac yb - \frac zx \right)^2=4 \\ & \Leftrightarrow \frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{x^2}+ 2 \cdot \left( \frac{yz}{bc}- \frac{xy}{ab}- \frac{zx}{ac} \right)=4 \\ & \Leftrightarrow \frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{x^2}+ 2 \cdot \dfrac{abc(ayz-cxy-bzx)}{a^2b^2c^2}=4 \\ & \Leftrightarrow \frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{x^2}+ 2 \cdot \dfrac{ayz-cxy-bzx}{abc}=4 \end{aligned}$$
Lại có $\frac ax- \frac by - \frac cz =0 \Leftrightarrow \frac{ayz-bzx-cxy}{xyz}=0 \Leftrightarrow ayz-bzx-cxy=0$. Do đó $$\frac{x^2}{a^2}+ \frac{y^2}{b^2}+ \frac{z^2}{x^2}= \boxed{4}.$$
- yellow likes this
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
#3
Posted 27-12-2012 - 16:34
Zaraki
-
- Phó Quản lý Toán Cao cấp
-
- 4273 posts
PQT
Tổng quát: Cho $\frac{a_1}{b_1}+ \frac{a_2}{b_2}+ \cdots + \frac{a_n}{b_n}=k$ với $n \in \mathbb{N}^*$. và $\frac{b_1}{a_1}+ \frac{b_2}{a_2}+ \cdots + \frac{b_n}{a_n}=0$. Chứng minh $$\frac{a_1^2}{b_1^2}+ \frac{a_2^2}{b_2^2}+ \cdots + \frac{a_n^2}{b_n^2}=k^2$$
Mọi người thử chứng minh coi có đúng không nhỉ
Mọi người thử chứng minh coi có đúng không nhỉ
Edited by Phạm Quang Toàn, 27-12-2012 - 16:34.
- yellow likes this
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
