Với 0 < x < 1 ; ta có : \(A=\dfrac{3}{1-x}+\dfrac{4}{x}\ge\dfrac{\left(\sqrt{3}+2\right)^2}{1-x+x}=\left(\sqrt{3}+2\right)^2\)

" = " \(\Leftrightarrow\dfrac{\sqrt{3}}{1-x}=\dfrac{2}{x}\)  \(\Leftrightarrow x=\dfrac{2}{\sqrt{3}+2}=4-2\sqrt{3}\) (t/m) 

Với a;b > 0 ; AD BĐT Cô-si ; ta được : \(a^4+1+1+1\ge4a\)  \(\Rightarrow a^4+3\ge4a\)  

\(\Rightarrow\dfrac{1}{a^4+4}\le\dfrac{1}{4a+1}\) . CMTT : \(\dfrac{1}{b^4+4}\le\dfrac{1}{4b+1}\)  

Suy ra : \(A=\dfrac{1}{a^4+4}+\dfrac{1}{b^4+4}\le\dfrac{1}{4a+1}+\dfrac{1}{4b+1}\)

Mặt khác : \(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+1\ge\dfrac{\left(1+1+1+1+1\right)^2}{4a+1}\)  ( B.C.S)

\(\Rightarrow\dfrac{4}{a}+1\ge\dfrac{25}{4a+1}\Rightarrow\dfrac{1}{25}\left(\dfrac{4}{a}+1\right)\ge\dfrac{1}{4a+1}\)

CMTT : \(\dfrac{1}{4b+1}\le\dfrac{1}{25}\left(\dfrac{4}{b}+1\right)\)

Suy ra : \(A\le\dfrac{4}{25}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{2}{25}=\dfrac{4}{25}.2+\dfrac{2}{25}=\dfrac{2}{5}\)

" = " \(\Leftrightarrow a=b=1\)

P/t \(\Leftrightarrow2cos2x.cosx-2.sin2x.cos2x=0\)

\(\Leftrightarrow cos2x\left(cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=1\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\x=2k\pi\end{matrix}\right.\)  ( k \(\in Z\) )

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=2k\pi\end{matrix}\right.\)

Từ GT ; ta có : \(0\le a;b;c\le1\)  \(\Rightarrow a\ge a^2;b\ge b^2;c\ge c^2\) 

Ta có : \(\sqrt{1+3a}\ge\sqrt{1+2a+a^2}=\sqrt{\left(a+1\right)^2}=\left|a+1\right|=a+1\)

CMTT : \(\sqrt{1+3b}\ge b+1;\sqrt{1+3c}\ge c+1\)

Suy ra : \(P\ge a+b+c+3=1+3=4\)

" = " \(\Leftrightarrow\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị 

Ta có : \(y=x^3-\left(m+6\right)x^2+\left(2m+9\right)x-2\left(C_m\right)\)

PTHĐGĐ của \(C_m\) với Ox : \(x^3-\left(m+6\right)x^2+\left(2m+9\right)x-2=0\)  (1) 

\(\Leftrightarrow\left(x-2\right)\left[x^2-\left(m+4\right)x+1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^2-\left(m+4\right)x+1=0\left(2\right)\end{matrix}\right.\)

(Cm) có 2 điểm cực trị nằm về 2 phía của Ox \(\Leftrightarrow\) (1) có 3 no p/b 

\(\Leftrightarrow\left(2\right)\) có 2 no p/b \(\ne2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(m+4\right)^2-4>0\\g\left(2\right)\ne0\end{matrix}\right.\)   \(\Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)\left(m+6\right)>0\\2^2-2\left(m+4\right)+1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< -6\\m>-2\end{matrix}\right.\\m\ne-\dfrac{3}{2}\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}m\ne-\dfrac{3}{2}\\m>-2\end{matrix}\right.\)

Vậy ... 

\(\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2=12\)

\(\Leftrightarrow\left(x+y+z\right)^2+x^2+y^2+z^2=12\)

\(\Leftrightarrow x^2+y^2+z^2=3\)

Dễ dàng c/m :  \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\)   \(\Rightarrow x^2+y^2+z^2\ge\dfrac{3^2}{3}=3\)

" = " \(\Leftrightarrow x=y=z=1\)

Khi đó : \(A=1^{2019}.3=3\)

Vậy ... 

E là trọng tâm \(\Delta ABC\Rightarrow\dfrac{ME}{MB}=\dfrac{1}{3}\)

F là trọng tâm \(\Delta ACD\Rightarrow\dfrac{MF}{MD}=\dfrac{1}{3}\)

Suy ra : \(\dfrac{ME}{MB}=\dfrac{MF}{MD}=\dfrac{EF}{BD}=\dfrac{1}{3}\)

ĐKXĐ : \(x\ge-1\)

\(\sqrt{x+1}+\sqrt{x+6}=5\)  \(\Leftrightarrow\sqrt{x+1}-2+\sqrt{x+6}-3=0\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{x+1}+2}+\dfrac{x-3}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\dfrac{1}{\sqrt{x+1}+2}+\dfrac{1}{\sqrt{x+6}+3}=0\left(PTVN\right)\end{matrix}\right.\)

Vậy ... 

\(\dfrac{1}{cos^2x}=\dfrac{sin^2x+cos^2x}{cos^2x}=1+tan^2x\left(đpcm\right)\)

Chu vi mảnh vườn : \(\left(12+\dfrac{42.2}{6}\right).2=52\left(m\right)\)

Đ/s : ...