Ta có: \(4x^2+4z^2=17\Rightarrow x^2+z^2=\frac{17}{4}\); \(4y\left(x+2\right)=5\Leftrightarrow2xy+4y=\frac{5}{2}\); \(20y^2+27=-16z\Rightarrow5y^2+4z=-\frac{27}{4}\)

\(\Rightarrow x^2+z^2-2xy-4y+5y^2+4z=-5\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(z^2+4z+4\right)+\left(4y^2-4y+1\right)=0\)\(\Leftrightarrow\left(x-y\right)^2+\left(z+2\right)^2+\left(2y-1\right)^2=0\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=-2\end{cases}}\)

\(\Rightarrow M=10.\frac{1}{2}+4.\frac{1}{2}+2019.\left(-2\right)=-4031\)

\(\sqrt{4x+2\sqrt{x}+1}\le\sqrt{4x+\dfrac{1}{2}\left(2^2+x\right)+1}=\sqrt{\dfrac{9x}{2}+3}\)

\(=\dfrac{1}{\sqrt{21}}.\sqrt{21}.\sqrt{\dfrac{9x}{2}+3}\le\dfrac{1}{2\sqrt{21}}\left(21+\dfrac{9x}{2}+3\right)=\dfrac{1}{2\sqrt{21}}\left(\dfrac{9x}{2}+24\right)\)

Tương tự và cộng lại:

\(A\le\dfrac{1}{2\sqrt{21}}\left(\dfrac{9}{2}\left(x+y+z\right)+72\right)=3\sqrt{21}\)

\(A_{max}=3\sqrt{21}\) khi \(x=y=z=4\)

\(A=1\sqrt{4x+2\sqrt{x}+1}+1.\sqrt{4y+2\sqrt{y}+1}+1\sqrt{4z+2\sqrt{z}+1}\)

\(\le\sqrt{\left(1+1+1\right)\left(4\left(x+y+z\right)+2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\right)}\)

\(=\sqrt{3.\left[51+\dfrac{4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2}\right]}\)

\(\le\sqrt{3.\left[51+\dfrac{x+y+z+12}{2}\right]}\)

\(=\sqrt{189}\)

Dấu "=" xảy ra <=> x = y = z = 4

  Ta có 1/x+1/y+1/z=0 
=>1/x+1/y=-1/z 
=>(1/x+1/y)^3= (-1/z)^3 
=>1/x^3+1/y^3+3.1/x.1/y.(1/x+1/y) =-1/z^3 
=>1/x^3+1/y^3+1/z^3= -3.1/x.1/y.(1/x+1/y) =3/(xyz) (vì 1/x+1/y=-1/z) 
Mặt khác: 1/x+1/y+1/z=0 
=>(xy+yz+zx)/(xyz)=0 
=>xy+yz+zx=0 
A=yz/x^2 +2yz + xz/y^2+ 2xz + xy/z^2+ 2 xy 
=xyz/x^3+xyz/y^3+xyz/z^3 +2(xy+yz+zx) (vì x,y,z khác 0) 
=xyz(1/x^3+1/y^3+1/z^3) (vì xy+yz+zx=0) 
=xyz.3/(xyz) (vì 1/x^3+1/y^3+1/z^3=3/(xyz) ) 
=3 
Vậy A=3.

\(Q=\left(x-3\right)\left(4x+5\right)+2019\)

\(=4x^2-7x-15+2019\)

\(=4x^2-7x+2004\)

\(=\left(2x-\frac{7}{4}\right)^2+\frac{32015}{16}\ge\frac{32015}{16}\forall x\)

Dấu "=" xảy ra<=>\(\left(2x-\frac{7}{4}\right)^2=0\Leftrightarrow2x=\frac{7}{4}\Leftrightarrow x=\frac{7}{8}\)

Giúp mk phần 2 vs m.n ơi

x² + 2y² + z² - 2xy - 2y - 4z + 5 = 0

<=> ( x² - 2xy + y² ) + ( y² - 2y + 1 ) + ( z² - 4z + 4 ) = 0

<=> ( x - y )² + ( y - 1 )² + ( z - 2 )² = 0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )² + ( y - 1 )² + ( z - 2 )²\(\ge\)0\(\forall\)x ; y ; z

Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )

Thay ( 1 ) vào A , ta được :

\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)

Vậy A = 2

Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)

Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)

Do \(x;y\in\left[0;2\right]\Rightarrow\left\{{}\begin{matrix}x\left(2-x\right)\ge0\\y\left(2-y\right)\ge0\end{matrix}\right.\) \(\Rightarrow2x^2+4y^2\le4x+8y\)

\(P\le3^0+5^0+3^z+4\left(x+2y\right)=2+3^z+4\left(6-z\right)=3^z-4z+26\)

Xét hàm \(f\left(z\right)=3^z-4z+26\) trên \(\left[0;2\right]\)

\(f'\left(z\right)=3^z.ln3-4=0\Rightarrow z=log_3\left(\dfrac{4}{ln3}\right)=a\)

\(f\left(0\right)=27\) ; \(f\left(2\right)=27\); \(f\left(a\right)\approx-1,1\)

\(\Rightarrow f\left(z\right)\le27\Rightarrow maxP=27\)

(Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(0;2;2\right)\))

Ồ mà khoan, bài trước bị nhầm lẫn ở chỗ \(3^{2x-x^2}+5^{2y-y^2}\ge3^0+5^0\) mới đúng, ko để ý bị ngược dấu đoạn này

Vậy giải cách khác:

\(0\le x;y;z\le2\Rightarrow x\left(2-x\right)\ge0\Rightarrow2x-x^2\ge0\)

Lại có: \(2x-x^2=1-\left(x-1\right)^2\le1\)

\(\Rightarrow0\le2x-x^2\le1\)

Tương tự ta có: \(0\le2y-y^2\le1\)

Xét hàm: \(f\left(t\right)=3^t-2t\) trên \(\left[0;1\right]\)

\(f'\left(t\right)=3^t.ln3-2=0\Rightarrow t=log_3\left(\dfrac{2}{ln3}\right)=a\)

\(f\left(0\right)=1;\) \(f\left(1\right)=1\) ; \(f\left(a\right)\approx0,73\)

\(\Rightarrow f\left(t\right)\le1\Rightarrow3^t-2t\le1\Rightarrow3^t\le2t+1\)

\(\Rightarrow3^{2x-x^2}\le2\left(2x-x^2\right)+1\)

Hoàn toàn tương tự, ta chứng minh được: 

\(5^t\le4t+1\) với \(t\in\left[0;1\right]\Rightarrow5^{2y-y^2}\le4\left(2y-y^2\right)+1\)

\(3^t\le4t+1\) với \(t\in\left[0;2\right]\Rightarrow3^z\le4z+1\)

\(\Rightarrow P\le2\left(2x-x^2\right)+4\left(2y-y^2\right)+4z+3+2x^2+4y^2=4\left(x+2y+z\right)+3=27\)

Lần này thì ko sai được rồi