Chủ đề này có 3 trả lời

[DOTOANNANG]

Tìm $\Delta\,\it{DEF}$ sao cho với bất kì $\Delta\,\it{ABC}$$:$

$$\it{DE}= \it{ab}\sqrt{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{C}}{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}}\,\,,$$

$$\it{EF}= \it{bc}\sqrt{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{A}}{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}}\,\,,$$

$$\it{FD}= \it{ca}\sqrt{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{B}}{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}}\,\,.$$

[DOTOANNANG]

Tìm $\Delta\,\it{DEF}$ sao cho với bất kì $\Delta\,\it{ABC}$$:$

$$\it{DE}= \it{ab}\sqrt{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{C}}{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}}\,\,,$$

$$\it{EF}= \it{bc}\sqrt{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{A}}{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}}\,\,,$$

$$\it{FD}= \it{ca}\sqrt{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{B}}{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}}\,\,.$$

Khi đó chứng minh rằng$:$

$$\frac{\it{3}}{\it{4}}\geqq \frac{\it{EF}\,.\,\it{FD}}{\it{bc}}+ \frac{\it{FD}\,.\,\it{DE}}{\it{ca}}+ \frac{\it{DE}\,.\,\it{EF}}{\it{ab}}$$

[DOTOANNANG]

Tìm $\Delta\,\it{DEF}$ sao cho với bất kì $\Delta\,\it{ABC}$$:$

$$\it{DE}= \it{ab}\sqrt{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{C}}{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}}\,\,,$$

$$\it{EF}= \it{bc}\sqrt{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{A}}{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}}\,\,,$$

$$\it{FD}= \it{ca}\sqrt{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{B}}{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}}\,\,.$$

Khi đó chứng minh rằng$:$

$$\frac{\it{b}+ \it{c}}{\it{EF}}\geqq \it{2}+ \frac{\it{b}+ \it{c}}{\it{a}}$$

[DOTOANNANG]

Tìm $\Delta\,\it{DEF}$ sao cho với bất kì $\Delta\,\it{ABC}$$:$

$$\it{DE}= \it{ab}\sqrt{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{C}}{\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}}\,\,,$$

$$\it{EF}= \it{bc}\sqrt{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{c}+ \it{a}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{A}}{\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}}\,\,,$$

$$\it{FD}= \it{ca}\sqrt{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}^{\,-\,\it{2}}+ \it{(}\,\,\it{a}+ \it{b}\,\,\it{)}^{\,-\,\it{2}}- \frac{\it{2}\,\cos\,\it{B}}{\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}}\,\,.$$

Khi đó chứng minh rằng$:$

$$\frac{\it{3}\,\it{abc}}{\it{4}\,\it{EF}\,.\,\it{FD}\,.\,\it{DE}}\geqq \frac{\it{a}}{\it{EF}}+ \frac{\it{b}}{\it{FD}}+ \frac{\it{c}}{\it{DE}}$$