MathematicsLover

a) $\Delta FOE\sim \Delta FEB \Rightarrow \frac{OF}{OE}=\frac{EF}{BE}=\frac{1}{2}$

$\frac{CK}{CE}=\frac{OF}{OE}=\frac{1}{2}\Rightarrow CK=\frac{1}{2}CE=EF$

b) Theo hệ thức lượng: $\frac{1}{OE^2}=\frac{1}{BE^2}+\frac{1}{EF^2}=\frac{1}{CE^2}+\frac{4}{CE^2}=\frac{5}{CE^2}\Rightarrow \frac{OE^2}{CE^2}=\frac{1}{5}$

$\frac{SOEF}{SECK}=\frac{1}{5}\Rightarrow \frac{SOECK}{SECK}=\frac{4}{5}$

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   Tìm tất cả các cặp số nguyên ( x;y) thoả mãn :

             

       $10x^{2}+50y^{2}+42xy+14x-6y+57<0$

$\Leftrightarrow 100x^2+500y^2+420xy+140x-60y+570<0$

$\Leftrightarrow [100x^2+20x(21y+7)+(21y+y)^2]-441y^2-294y-49+500y^2-60y+570<0$

$\Leftrightarrow (10x+21y+7)^2+59y^2-354y+521<0$

$\Leftrightarrow (10x+21y+7)^2+59(y-3)^2<10$

$\Rightarrow 59(y-3)^2<10\Rightarrow (y-3)^2=0\Leftrightarrow y=3$

$\Rightarrow (10x+70)^2<10\Rightarrow 100(x+7)^2<10\Rightarrow (x+7)^2=0\Leftrightarrow x=-7$

Vậy $(x;y)=(-7;3)$

2. $(sinP+cosP)^{2}\leq 2(sin^2P+cos^2P)=2\Rightarrow MaxP = \sqrt{2} \Leftrightarrow sinP+cosP=\frac{\sqrt{2}}{2}\Leftrightarrow P=45^{\circ}$

4,

Áp dụng BĐT $\frac{1}{x+y}\leq \frac{1}{4}(\frac{1}{x}+\frac{1}{y}) \forall x,y> 0$

Ta có:$A=\frac{ab}{c+12}+\frac{bc}{a+12}+\frac{ca}{b+12}= \frac{ab}{(a+c)+(b+c)}+\frac{bc}{(a+b)+(a+c)}+\frac{ca}{(a+b)+(b+c)}$

$A\leq \frac{ab}{4}(\frac{1}{a+c}+\frac{1}{b+c})+\frac{bc}{4}(\frac{1}{a+b}+\frac{1}{a+c})+\frac{ac}{4}(\frac{1}{a+b}+\frac{1}{b+c})$

$A\leq \frac{1}{4}(a+b+c)=3$

Dấu "=" xảy ra khi và chỉ khi a=b=c=1

5,

$a+b\geq 2\sqrt{ab}=>12\geq (a+b)^3+4ab\geq 8(\sqrt{ab})^{3}+4(\sqrt{ab})^{2}$

$\Rightarrow (\sqrt{ab}-1)(2ab+2\sqrt{ab}+3)\leq 0\Rightarrow \sqrt{ab}\leq 1$

$\Rightarrow (\sqrt{ab}-1)(\sqrt{a}-\sqrt{b})^2\leq 0$

$\Leftrightarrow (a+b+2)(1+\sqrt{ab})\leq 2(a+1)(b+1)$

$\Leftrightarrow \frac{1}{a+1}+\frac{1}{b+1}\leq \frac{2}{1+\sqrt{ab}}$

Cần chứng minh: $\frac{2}{1+\sqrt{ab}} +2015ab\leq 2016$

Đặt $\sqrt{ab}=t(0\leq t\leq 1)$

$\frac{2}{t+1}+2015t^2\leq 2016\Leftrightarrow 2015t^3+2015t^2-2016t-2014\leq 0$

$\Leftrightarrow (t-1)(2015t^2+4030t+2014)\leq 0$ (luôn đúng)

Vậy $\frac{1}{1+a}+\frac{1}{1+b}+2015ab\leq 2016$

Dấu "=" xảy ra <=> a=b=1