Chủ đề này có 92 trả lời

[25 minutes]

Góp vui chút. Cho các số thực $x,y,z$ thỏa mãn $\left\{\begin{matrix}x,y,z \epsilon [1;9] &  & \\ x\geq y,x\geq z &  & \end{matrix}\right.$

Tìm GTLN và GTNN của $P=\frac{x}{x+2y}+\frac{y}{y+z}+\frac{z}{z+x}$

Ta có $P=\frac{1}{1+\frac{2y}{x}}+\frac{1}{1+\frac{z}{y}}+\frac{1}{1+\frac{x}{z}}\geqslant \frac{1}{1+\frac{2y}{x}}+\frac{2}{1+\sqrt{\frac{x}{y}}}$

Đặt $t=\sqrt{\frac{x}{y}}\Rightarrow t \in \left [ 1;3 \right ]$, do $1 \leqslant y \leqslant x \leqslant 9$

Khi đó $P\geqslant f(t)=\frac{1}{1+\frac{2}{t^2}}+\frac{2}{1+t}=\frac{t^2}{t^2+2}+\frac{2}{t+1}$

$\Rightarrow f'(t)=\frac{4t}{(t^2+2)^2}-\frac{2}{(t+1)^2}=0\Leftrightarrow t=2,t=\sqrt[3]{2}$

Lập bảng biến thiên ta có được 

    $P\geqslant f(t)\geqslant \frac{4}{3}\Leftrightarrow (a,b,c)=(4,1,2)$ ứng với $t=2$

P/S: Bài này không có max

[dhdhn]

Làm như bạn chỉ tìm được min với đk t $ <= $ 1 thôi chưa giải quyết được max

Do x²+y²=1+xy nên (x+y)²=1+3xy 

Từ đó nên xy $>=$ -1/3

[dhdhn]

Làm như bạn chỉ tìm được min với đk t $ <= $ 1 thôi chưa giải quyết được max

$x^{2}+y^{2}=1+xy \Rightarrow \left\{\begin{matrix}1+xy\geq 2xy & & \\ (x+y)^{2}=1+3xy\geq 0 & & \end{matrix}\right. \Rightarrow \frac{-1}{3}\leq t\leq 1$

[phamxuanquynhtb98]

Mọi người giúp em bài này:Cho a, b, c không âm, a+b+c=1 chứng minh rằng: (ab)³+(bc)³+(ca)³ <=1/64

[Lam Ba Thinh]

$$P=\frac{a}{b^{2}+c^{2}}+\frac{b}{a^{2}+c^{2}}+\frac{c}{a^{2}+b^{2}}=\frac{a^2}{a(1-a^2)}+\frac{b^2}{b(1-b^2)}+\frac{c^2}{c(1-c^2)}$$
Xét hàm số $f(x)=x(1-x^2)$ với $x>0$
Ta có: $f'(x) =1-3x^2 ; f'(x) =0 \Leftrightarrow x=\frac{1}{\sqrt{3}}>0$
Từ bảng biến thiên ta có $f(x)\le \frac{2}{3\sqrt{3}}\,\,\forall x>0$
Khi đó $$P =\frac{a^2}{f(a)}+\frac{b^2}{f(b)}+\frac{c^2}{f( c)}\ge \frac{3\sqrt{3}}{2}(a^2+b^2+c^2)=\frac{3\sqrt{3}}{2}$$
Dấu "=" xảy ra khi và chỉ khi $a=b=c=\frac{1}{\sqrt{3}}$

Tìm max?

[Dung Du Duong]

Giúp e bài này với:1. Cho tam giác ABC nhọn, tìm MIN: P = tanA + 2tanB + 3tanC

                           2. Cho a,b > 0 và a+b+2$\sqrt{(a+2)(b+2)}$=12. Tìm MIN: A=$\frac{a^{3}}{b+2}+\frac{b^{3}}{a+2}+\frac{48}{a+b}$   

Bài viết đã được chỉnh sửa nội dung bởi Dung Du Duong: 31-07-2015 - 21:15

[kimchitwinkle]

Cho x,y thay đổi thỏa mãn : $x^{2}-xy+y^{2}=1$ . Tìm max và min của $P=\frac{\left ( x^{2} -1\right )^{2}+(y^{2}-1)^{2}+2xy(xy-1)+3}{x^{2}+y^{2}-3}$

[kudoshinichihv99]

Cho x,y thay đổi thỏa mãn : $x^{2}-xy+y^{2}=1$ . Tìm max và min của $P=\frac{\left ( x^{2} -1\right )^{2}+(y^{2}-1)^{2}+2xy(xy-1)+3}{x^{2}+y^{2}-3}$

Ta có $P=\frac{x^4+y^4-2(x^2+y^2)+2+2x^2y^2-2xy+3}{xy-2}=\frac{(x^2+y^2)^2-2(xy+1)-2xy+5}{xy-2}=\frac{x^2y^2-2xy+4}{xy-2}$

Xét hàm $f(t)=\frac{t^2-2t+4}{t-2},với t=xy=x^2+y^2-1\geq 2xy-1<=>t\leq 1$,..

[HocLop]

Chứng minh rằng với mọi x ta đều có: cos$^2$(x - a) + sin$^2$(x - b) - 2cos(x - a).sin(x - b).sin(a - b) = cos$^2$(a - b).

[duyen040502]

cho góp ý bài 7 nhế: bạn thử x=1 thỏa $x> 0$ thì $\sin x< x-\frac{x^{3}}{6}$,vậy bất đẳng thức của bạn sai nhé!

ủa đúng mà bạn, sin(1) = 0,84......; 1-$\frac{1^{3}}{6}$ = 0,83......

[anny anh]

Cho a,b,c thực không âm t/m x+y+z=1

Tim Max P=$\frac{x-2}{3-y-z}+\frac{y-2}{3-x-z}+\frac{z-2}{3-x-y}+2x^2+2y^2+2z^2+xy+yz+xz$

[DOTOANNANG]

$\it{2}$ bài được giải bằng đạo hàm trên $\lceil$ Diễn đàn Toán học $\rfloor$$:$

 

$\lceil$ https://diendantoanh...e-1#entry719851 $\rfloor$

 

 

$\lceil$ https://diendantoanh...e-1#entry719749 $\rfloor$

 

Từ $\it{:}$ $\it{32}\,\it{x}^{\,\it{6}}+ \it{4}\,\it{y}^{\,\it{3}}= \it{1}\,\,\Rightarrow \,\,\it{(}\,\,\it{y}- \it{2}\,\it{x}^{\,\it{2}}\,\,\it{)}\it{(}\,\,\it{x}- \frac{\it{1}}{\it{2}}\,\,\it{)}\leqq \it{0}\,\,,\,\,\it{(}\,\,\it{x}- \frac{\it{1}}{\it{2}}\,\,\it{)}\it{(}\,\,\it{y}- \frac{\it{1}}{\it{2}}\,\,\it{)}\leqq \it{0}\,\,,\,\,{\it{y}}'= -\,\frac{\it{16}\,\it{x}^{\,\it{5}}}{\it{y}^{\,\it{2}}}$ $\it{.}$ Khi đó $\it{:}$
${\it{P}}'\it{(}\,\,\it{x}\,\,\it{)}=$ $\frac{{\it{[}\,\,\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{5}}\,\,\it{]}}'\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}- {\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}}'\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{5}}}{\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}^{\,\it{2}}}$ $= $

$=$ $\frac{\it{5}\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{4}}\left ( \it{4}\,\it{x}- \it{16}\,\frac{\it{x}^{\,\it{5}}}{\it{y}^{\,\it{2}}} \right )\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}- \left ( \it{6}\,\it{x}- \it{3}- \it{96}\,\frac{\it{x}^{\,\it{5}}}{\it{y}}+ \it{48}\,\frac{\it{x}^{\,\it{5}}}{\it{y}^{\,\it{2}}} \right )\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{5}}}{\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}^{\,\it{2}}}$ $= $

$=$ $\frac{\it{(}\,\,\it{2}\,\it{x}^{\,\it{2}}+ \it{y}+ \it{3}\,\,\it{)}^{\,\it{4}}\it{\{}\,\,\it{20}\,\it{x}\it{(}\,\,\it{y}- \it{2}\,\it{x}^{\,\it{2}}\,\,\it{)}\it{(}\,\,\it{y}+ \it{2}\,\it{x}^{\,\it{2}}\,\,\it{)}\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}- \it{3}\it{[}\,\,\it{x}^{\,\it{5}}\it{(}\,\,\it{16}- \it{32}\,\it{y}\,\,\it{)}+ \it{y}^{\,\it{2}}\it{(}\,\,\it{2}\,\it{x}- \it{1}\,\,\it{)}\,\,\it{]}\,\,\it{\}}}{\it{y}^{\,\it{2}}\it{[}\,\,\it{3}\it{(}\,\,\it{x}^{\,\it{2}}+ \it{y}^{\,\it{2}}\,\,\it{)}- \it{3}\it{(}\,\,\it{x}+ \it{y}\,\,\it{)}+ \it{2}\,\,\it{]}^{\,\it{2}}}$

Dễ thấy ngay $\it{:}$ $\it{(}\,\,\it{x}- \frac{\it{1}}{\it{2}}\,\,\it{)}{\it{P}}'\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{0}\,\,\Rightarrow \,\,\it{P}\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{P}\it{(}\,\,\frac{\it{1}}{\it{2}}\,\,\it{)}= \it{2048}$




 

$\lceil$ Ý nghĩa hình học của đạo hàm $\it{(}$ $\it{!}$ $\it{)}$   $\rfloor$

By H-a-i-D-a-n-g-e-l $\it{(}$ $\it{D-...-A-N-G}$ $\it{)}$

 

Từ $\it{:}$ $\it{x}^{\,\it{2}}+ \it{2}\,\it{y}^{\,\it{2}}= \frac{\it{8}}{\it{3}}\,\,\Rightarrow \,\,\it{(}\,\,\it{x}- \it{2}\,\it{y}\,\,\it{)}\it{(}\,\,\it{x}- \frac{\it{4}}{\it{3}}\,\,\it{)}\geqq \it{0}\,\,,\,\,\it{2}\,{\it{y}}'= -\,\frac{\it{x}}{\it{y}}$ $\it{.}$ Đặt $\it{:}$ $\it{f}\,\it{(}\,\,\it{x}\,\,\it{)}= \it{7}\,\it{(}\,\,\it{x}+ \it{2}\,\it{y}\,\,\it{)}- \it{4}\,\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}$ $\it{.}$ Khi đó $\it{:}$

${\it{f}}'\,\it{(}\,\,\it{x}\,\,\it{)}= \it{7}\,\it{(}\,\,\it{1}+ \it{2}\,{\it{y}}'\,\,\it{)}- \frac{\it{4}\,\it{(}\,\,\it{2}\,\it{x}+ \it{2}\,\it{y}+ \it{2}\,\it{x}{\it{y}}'+ \it{16}\,\it{y}{\it{y}}'\,\,\it{)}}{\it{2}\,\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}}=$

$= \it{7}\,\it{(}\,\,\it{1}- \frac{\it{x}}{\it{y}}\,\,\it{)}- \frac{\it{2}\,\it{(}\,\,\it{2}\,\it{x}+ \it{2}\,\it{y}- \frac{\it{x}^{\,\it{2}}}{\it{y}}- \it{8}\,\it{x}\,\,\it{)}}{\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}}=$

$= \it{7}\,\it{(}\,\,\it{2}- \frac{\it{x}}{\it{y}}\,\,\it{)}+ \frac{\it{2}\,\it{(}\,\,\it{6}\,\it{xy}- \it{2}\,\it{y}^{\,\it{2}}+ \it{x}^{\,\it{2}}\,\,\it{)}}{\it{y}\,\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}}- \it{7}=$

$= \frac{\it{1}}{\it{y}}\,\left \{ -\,\it{7}\,\it{(}\,\,\it{x}- \it{2}\,\it{y}\,\,\it{)}+ \frac{\it{4}\,\it{(}\,\it{6}\,\it{xy}- \it{2}\,\it{y}^{\,\it{2}}+ \it{x}^{\,\it{2}}\,\,\it{)}^{\,\it{2}}- \it{49}\,\it{y}^{\,\it{2}}\,\it{(}\,\,\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}\,\,\it{)}^{\,\it{2}}}{\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}\left [ \it{2}\,\it{(}\,\it{6}\,\it{xy}- \it{2}\,\it{y}^{\,\it{2}}+ \it{x}^{\,\it{2}}\,\,\it{)}+ \it{7}\,\it{y}\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}} \right ]} \right \}=$

$= -\,\frac{\it{x}- \it{\it{2}\,\it{y}}}{\it{y}}\,\left \{ \,\,\underbrace{\it{7}- \frac{\it{4}\,\it{x}^{\,\it{3}}+ \it{56}\,\it{x}^{\,\it{2}}\it{y}+ \it{191}\,\it{xy}^{\,\it{2}}+ \it{188}\,\it{y}^{\,\it{3}}}{\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}}\left [ \it{2}\,\it{(}\,\it{6}\,\it{xy}- \it{2}\,\it{y}^{\,\it{2}}+ \it{x}^{\,\it{2}}\,\,\it{)}+ \it{7}\,\it{y}\sqrt{\it{x}^{\,\it{2}}+ \it{2}\,\it{xy}+ \it{8}\,\it{y}^{\,\it{2}}} \right ]}}_{> \it{0}}\,\, \right \}= $

Ta sẽ chứng minh bất đẳng thức trong dấu ngoặc nhọn luôn đúng $\it{(}\,\,\it{!}\,\,\it{)}$ Thật vậy $\it{,}$ với $\it{t}= -\,\it{2}\,{\it{y}}'> 0$ thì $\it{:}$

$\frac{\mathrm{d} }{\mathrm{d} \it{t}}\left ( \,\,\underbrace{\it{2}\,\it{(}\,\,\it{6}\,\it{t}- \it{2}+ \it{t}^{\,\it{2}}\,\,\it{)}+ \it{7}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}\,\,\it{)}}_{> \it{0}\,\,\Leftarrow \,\,\it{(}\,\,\star\,\,\it{)}}\,\, \right )= \frac{\it{14}\,\it{(}\,\,\it{t}+ \it{1}\,\,\it{)}}{\it{2}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}}+ \it{4}\,\it{(}\,\,\it{t}+ \it{3}\,\,\it{)}> \it{0}\,\,\it{(}\,\,\star\,\,\it{)}$

$\it{(}$ Hay $\it{)}$ tương đương với $\it{:}$

$\it{14}\,\it{(}\,\,\it{6}\,\it{t}- \it{2}+ \it{t}^{\,\it{2}}\,\,\it{)}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}> \it{4}\,\it{t}^{\,\it{3}}+ \it{56}\,\it{t}^{\,\it{2}}+ \it{191}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}$

Xét $\it{0}< \it{t}\leqq \it{2}$ thì$\it{:}$

$\it{14}\,\it{(}\,\,\it{6}\,\it{t}- \it{2}+ \it{t}^{\,\it{2}}\,\,\it{)}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}\geqq \it{4}\,\it{t}^{\,\it{3}}+ \it{241}\,\it{t}^{\,\it{2}}+ \it{192}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}> \it{4}\,\it{t}^{\,\it{3}}+ \it{56}\,\it{t}^{\,\it{2}}+ \it{191}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}$

Xét $\it{t}\geqq \it{2}$ thì $\it{:}$

$\it{14}\,\it{(}\,\,\it{6}\,\it{t}- \it{2}+ \it{t}^{\,\it{2}}\,\,\it{)}\,\sqrt{\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}}\geqq \it{4}\,\it{t}^{\,\it{3}}+ \it{56}\,\it{t}^{\,\it{2}}+ \it{562}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}> \it{4}\,\it{t}^{\,\it{3}}+ \it{56}\,\it{t}^{\,\it{2}}+ \it{191}\,\it{t}+ \it{188}- \it{49}\,\it{(}\,\,\it{t}^{\,\it{2}}+ \it{2}\,\it{t}+ \it{8}\,\,\it{)}$

Ta được $\it{:}$ $\it{(}\,\,\it{x}- \it{2}\,\it{y}\,\,\it{)}{\it{f}}'\,\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{0}\,\,\Rightarrow\,\, \it{(}\,\,\it{x}- \frac{\it{4}}{\it{3}}\,\,\it{)}{\it{f}}'\,\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{0}\,\,\Rightarrow\,\, \it{f}\,\it{(}\,\,\it{x}\,\,\it{)}\leqq \it{f}\,\it{(}\,\,\frac{\it{4}}{\it{3}}\,\,\it{)}= \it{8}$

$\lceil$ Ý nghĩa hình học của đạo hàm $\it{(}$ $\it{!}$ $\it{)}$    $\rfloor$

By H-a-i-D-a-n-g-e-l $\it{(}$ $\it{D-...-A-N-G}$ $\it{)}$

 

 

[Andy Duong]

nhiều câu khó quá, cảm ơn mọi người đã hỗ trợ ^^