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Gọi O là tâm đường tròn \(\Rightarrow\) O là trung điểm BC
\(\stackrel\frown{BE}=\stackrel\frown{ED}=\stackrel\frown{DC}\Rightarrow\widehat{BOE}=\widehat{EOD}=\widehat{DOC}=\dfrac{180^0}{3}=60^0\)
Mà \(OD=OE=R\Rightarrow\Delta ODE\) đều
\(\Rightarrow ED=R\)
\(BN=NM=MC=\dfrac{2R}{3}\Rightarrow\dfrac{NM}{ED}=\dfrac{2}{3}\)
\(\stackrel\frown{BE}=\stackrel\frown{DC}\Rightarrow ED||BC\)
Áp dụng định lý talet:
\(\dfrac{AN}{AE}=\dfrac{MN}{ED}=\dfrac{2}{3}\Rightarrow\dfrac{EN}{AN}=\dfrac{1}{2}\)
\(\dfrac{ON}{BN}=\dfrac{OB-BN}{BN}=\dfrac{R-\dfrac{2R}{3}}{\dfrac{2R}{3}}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{EN}{AN}=\dfrac{ON}{BN}=\dfrac{1}{2}\) và \(\widehat{ENO}=\widehat{ANB}\) (đối đỉnh)
\(\Rightarrow\Delta ENO\sim ANB\left(c.g.c\right)\)
\(\Rightarrow\widehat{NBA}=\widehat{NOE}=60^0\)
Hoàn toàn tương tự, ta có \(\Delta MDO\sim\Delta MAC\Rightarrow\widehat{MCA}=\widehat{MOD}=60^0\)
\(\Rightarrow\Delta ABC\) đều
Đặt A = \(\dfrac{3}{\sqrt{x}+2}\)
Để A \(\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in z\Leftrightarrow\left(\sqrt{x}+2\right)\inƯ\left(3\right)\)
\(\Rightarrow\left(\sqrt{x}+2\right)\in\left\{\pm1;\pm3\right\}\)
Ta có bẳng sau:
| \(\sqrt{x}+2\) | -3 | -1 | 1 | 3 |
| \(\sqrt{x}\) | -5(loại) | -3(loại) | -1(loại) | 1 |
| \(x\) | loại | loại | loại | 1 |
Vậy \(x=1\) thỏa mãn
\(VT=\dfrac{xy\sqrt{z-2}+yz\sqrt{x-3}+xz\sqrt{y-4}}{xyz}\)
\(=\dfrac{\sqrt{z-2}}{z}+\dfrac{\sqrt{x-3}}{x}+\dfrac{\sqrt{y-4}}{y}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{\sqrt{z-2}}{z}=\dfrac{\sqrt{2\left(z-2\right)}}{\sqrt{2}z}\le\dfrac{\dfrac{2+z-2}{2}}{\sqrt{2}z}=\dfrac{1}{2\sqrt{2}}\)
\(\dfrac{\sqrt{x-3}}{x}=\dfrac{\sqrt{3\left(x-3\right)}}{\sqrt{3}x}\le\dfrac{\dfrac{3+x-3}{2}}{\sqrt{3}x}=\dfrac{1}{2\sqrt{3}}\)
\(\dfrac{\sqrt{y-4}}{y}=\dfrac{\sqrt{4\left(y-4\right)}}{\sqrt{4}y}\le\dfrac{\dfrac{4+y-4}{2}}{\sqrt{4}y}=\dfrac{1}{2\sqrt{4}}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT=\dfrac{\sqrt{z-2}}{z}+\dfrac{\sqrt{x-3}}{x}+\dfrac{\sqrt{y-4}}{y}\le\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+\dfrac{1}{2\sqrt{4}}=VP\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=6\\y=8\\z=4\end{matrix}\right.\)
Lời giải:
Ta có \(P=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{4ab}+\frac{1}{4ab}+4ab\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a^2+b^2}+\frac{1}{2ab}\geq \frac{4}{a^2+b^2+2ab}=\frac{4}{(a+b)^2}\geq 4\)
Áp dụng BĐT AM-GM: \(\frac{1}{4ab}+4ab\geq 2\).
Và \(1\geq a+b\geq 2\sqrt{ab}\rightarrow ab\leq \frac{1}{4}\)
Do đó \(P\geq 4+1+2=7\) hay \(P_{\min}=7\)
Dấu bằng xảy ra khi \(a=b=\frac{1}{2}\)
cảm ơn bn nhiều lắm 
b , Ta có : \(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}\) = \(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
d , Ta có : \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=\left(\dfrac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}\right)\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)= \(\dfrac{\left(1-a\right)+\sqrt{a}\left(1-a\right)}{1-\sqrt{a}}.\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
= \(\dfrac{\left(1-a\right)\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)\left(1-a\right)^2}\)
= \(\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}{\left(1-a\right)}=\dfrac{\left(1-a\right)}{\left(1-a\right)}=1\)
\(2\sqrt{3a}-\sqrt{75a}+\sqrt{\dfrac{13,5}{2a}}-\dfrac{2}{5}\sqrt{300a^3}\left(a>0\right)\)
=\(2\sqrt{3a}-\sqrt{5^2\cdot3a}+a\sqrt{\dfrac{13,5\cdot2a}{\left(2a\right)^2}}-\dfrac{2}{5}\sqrt{10^2\cdot a^2\cdot2}\)
=\(\left(2-5+\dfrac{3}{2}-4a\right)\sqrt{a}\)
=\(\dfrac{-11}{2}a\sqrt{a}\)
\(\sqrt{x-2}+\sqrt{4-x} = 2x^{2}-5x-1\) ( ĐKXĐ : \(2\le x\le4\) )