Ta có pt <=> \(2\sqrt{x-2}+2\sqrt{y+2009}+2\sqrt{z-2010}=x+y+z\)

<=> \(x-2-2\sqrt{x-2}+1+y+2009-2\sqrt{y+2009}+1+z-2010-2\sqrt{z-2010}+1=0\)

<=> \(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2009}-1\right)^2+\left(\sqrt{z-2010}-1\right)^2=0\)

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^_^

nát cả óc!

-kindle (sử dụng ở ứng dụng Android

Bạn xem lại đề câu b và c nhé !

a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)

\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)

\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ

\(\Rightarrow x\ge2\) thỏa mãn đề.

d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)

Pt tương đương :

\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )

e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)

\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)

Phương trình (1) tương đương :

\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)

\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )

b) đk: \(x>2012;y>2013\)

pt \(\frac{16}{\sqrt{x-2012}}+\sqrt{x-2012}+\frac{1}{\sqrt{y-2013}}+\sqrt{y-2013}=10\)

\(VT\ge2\sqrt{\frac{16}{\sqrt{x-2012}}.\sqrt{x-2012}}+2\sqrt{\frac{1}{\sqrt{y-2013}}.\sqrt{y-2013}}=8+2=10\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-2012=16\\y-2013=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2028\\y=2014\end{cases}}\)

Điều kiện : \(x\ge2;y\ge-2009;z\ge2010;x+y+z\ge0\)

PT <=> \(2.\sqrt{x-2}+2.\sqrt{y+2009}+2.\sqrt{z-2010}=x+y+z\)

Áp dụng B ĐT Cô- si với 2 số dương a; b : \(2\sqrt{ab}\le a+b\) ta có:

\(2.\sqrt{x-2}\le x-2+1=x-1\)

\(2.\sqrt{y+2009}\le y+2009+1=y+2010\)

\(2.\sqrt{z-1010}\le z-2010+1=z-2009\)

=> \(2.\sqrt{x-2}+2.\sqrt{y+2009}+2.\sqrt{z-2010}\le x-1+y+2010+z-2009=x+y+z\)

Dấu "=" xảy ra <=> x - 2 = 1 ; y + 2009 = 1; z - 2010 = 1

=> x = 3; y = -2008; z = 2011 là nghiệm của PT

Điều kiện \(x\ge2\) vs \(y\ge-2009\) vs \(z\ge2010\)  Khi đó

PT \(\Leftrightarrow\) \(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2009}-1\right)^2+\left(\sqrt{z-2010}-1\right)^2=0\)

nên => x=3 ; y=-2008 vs z=2011

\(x-2008=X;y-2009=Y;z-2010=Z\)

\(\sqrt{X}+\sqrt{Y}+\sqrt{Z}+3012=\frac{1}{2}\left(X+Y+Z+2008+2009+2010\right)\)

\(2.\sqrt{X}+2\sqrt{Y}+2\sqrt{Z}+2.3012=X+Y+Z+2009\cdot3\)

\(\left(X-2\sqrt{X}+1\right)+\left(Y-2\sqrt{Y}+1\right)+\left(Z-2\sqrt{Z}+1\right)+3.2008=2.3012\)

\(\left(\sqrt{X}-1\right)^2+\left(\sqrt{Y}-1\right)^2+\left(\sqrt{Z}-1\right)^2=2.3012-3.2008=0\)

\(X=1;Y=1;Z=1\Rightarrow x=2009;y=2010;z=2011\)

Làm hơi tắt , thông cảm  ;))

Từ (1) \(\Rightarrow36=\left(x+y+z\right)^2\Leftrightarrow36=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

          \(\Leftrightarrow36=18+2\left(xy+yz+zx\right)\Leftrightarrow xy+yz+zx=9\)(4)

Từ (3) \(\Rightarrow16=\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\Leftrightarrow16=x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

          \(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=5\Leftrightarrow\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2=25\)

         \(\Leftrightarrow xy+yz+zx+2\left(\sqrt{xy^2z}+\sqrt{xyz^2}+\sqrt{x^2yz}\right)=25\)

         \(\Leftrightarrow\sqrt{xyz}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)=8\Leftrightarrow\sqrt{xyz}=\frac{8}{4}\Leftrightarrow xyz=4\)(5)

Vậy hệ đã cho tương đương với :

\(\hept{\begin{cases}x+y+z=6\left(1\right)\\xy+yz+zx=9\left(4\right)\\xyz=4\left(5\right)\end{cases}}\)

Từ (5) \(\Rightarrow yz=\frac{4}{x}\)(Dễ thấy \(x,y,z>0\))

     (4)  \(\Leftrightarrow xy+yz+zx+x^2=9+x^2\Leftrightarrow x\left(x+y+z\right)+yz=9+x^2\)

           \(\Leftrightarrow x.6+\frac{4}{x}=9+x^2\Leftrightarrow x^3-6x^2+9x-4=0\)

           \(\Leftrightarrow\left(x-1\right)^2\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}.}\)

Thế vào ta suy ra hệ có các nghiệm : \(\left(x,y,z\right)=\left(1,1,4\right),\left(1,4,1\right),\left(4,1,1\right).\)

thanks bạn Đào Thu Hòa 

nhân cả 2 vế với 2 ta có

\(2\sqrt{x-2}+2\sqrt{y+2000}+2\sqrt{z-2001}=x+y+z\)

\(\left(x-2\right)-2\sqrt{x-2}+1+\left(y+2000\right)-2\sqrt{y+2000}+1+\left(z-2001\right)-2\sqrt{z-2001}+1=0\)

\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2000}-1\right)^2+\left(\sqrt{z-2001}-1\right)^2=0\)

cho cả 3 cái =0 thì giả ra x=3  y=-1999  z=2002

how about the technology in the future Which things will happen Draw a picture about the technology in the future Note You can draw everything but they are different from now Please help me