Thưa bn mk đã làm ra nhưng không biết có đúng không. Xem nhá:

Ta có:

\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2001}-1}{y-2001}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\Leftrightarrow"\frac{1}{\sqrt{x-2009}}-\frac{1}{2}"^2+\)

\("\frac{1}{\sqrt{y-2010}}-\frac{1}{2}"^2-"\frac{1}{\sqrt{z-2011}}-\frac{1}{2}"^2=0\)

\(\Rightarrow x=2013;y=2014;z=2015\)

P/s: Bn thay Ngoặc Kép thành Ngoặc Đơn nhé

ĐXKĐ: ...

Bình phương 2 vế:

\(x-y+2010=x+y+2010-2\sqrt{xy}+2\sqrt{2010x}-2\sqrt{2010y}\)

\(\Leftrightarrow y-\sqrt{xy}-\left(\sqrt{2010y}-\sqrt{2010x}\right)=0\)

\(\Leftrightarrow\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)-\sqrt{2010}\left(\sqrt{y}-\sqrt{x}\right)=0\)

\(\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{2010}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}y=x\\y=2010\end{matrix}\right.\)

Vậy nghiệm của pt là: \(\left[{}\begin{matrix}x=y\ge0\\\left\{{}\begin{matrix}y=2010\\x\ge0\end{matrix}\right.\end{matrix}\right.\)

\(ĐKXĐ:x\ne2009;y\ne2010;z\ne2011;x,y,z\in R\)

\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{x-2009}-\frac{\sqrt{x-2009}}{x-2009}+\frac{1}{y-2010}-\frac{\sqrt{y-2011}}{y-2011}+\frac{1}{z-2011}-\frac{\sqrt{z-2011}}{z-2011}=\frac{-3}{4}\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}^2}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{y-2010}^2}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{z-2011}^2}+\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^{^2}+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow\frac{1}{\sqrt{x-2009}}=\frac{1}{2};\frac{1}{\sqrt{y-2010}}=\frac{1}{2};\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\)

\(\Leftrightarrow x=2013;y=2014;z=2015\inĐKXĐ\)

  VẬY       \(x=2013;y=2014;z=2015\)

ko biet E=MC'2

tui cũng ko biết làm nữa

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

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

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

...

^_^

nát cả óc!

-kindle (sử dụng ở ứng dụng Android

(+) 2010>=x > y > 0 

=> \(\sqrt{x}+\sqrt{2010-y}>\sqrt{2010-x}+\sqrt{y}\left(loại\right)\)

(+) 0< x < y =< 2010

=> \(\sqrt{2010-x}+\sqrt{y}>\sqrt{2010-y}+\sqrt{x}\left(loại\right)\) 

(+) với x = y tm 

thay vào pt (1) giải pt 

Giải phưởng trình ra nhé

thắng ơi phải xét như vậy ak 

a) ĐK: \(x>2009;y>2010;z>2011\)

\(\Leftrightarrow\frac{\sqrt{x-2009}-1}{x-2009}-\frac{1}{4}+\frac{\sqrt{y-2010}-1}{y-2010}-\frac{1}{4}+\frac{\sqrt{z-2011}-1}{z-2011}-\frac{1}{4}=0\)

\(\Leftrightarrow\frac{-\left(\sqrt{x-2009}-2\right)^2}{4\left(x-2009\right)}+\frac{-\left(\sqrt{y-2010}-2\right)^2}{4\left(y-2010\right)}+\frac{-\left(\sqrt{z-2011}-2\right)^2}{4\left(z-2011\right)}=0\left(1\right)\)

Dễ thấy với đkxđ thì \(VT\left(1\right)\le0\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}\left(tm\right)}}\)

\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)(*)

\(ĐK:\orbr{\begin{cases}x\ge3\\x\le-3\end{cases}}\)

(*)\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\)

Xét phương trình\(\sqrt{x+3}+\sqrt{x-3}=0\)(**) có \(\sqrt{x+3}\ge0;\sqrt{x-3}\ge0\)nên (**) xảy ra khi \(\hept{\begin{cases}\sqrt{x+3}=0\\\sqrt{x-3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=3\end{cases}}\left(L\right)\)

Vậy phương trình có một nghiệm duy nhất là 3

a. ĐK : x > 2009 ; y > 2010 ; z > 2011 

Pt <=> \(\frac{1-\sqrt{x-2009}}{x-2009}+\frac{1-\sqrt{y-2010}}{y-2010}+\frac{1-\sqrt{z-2011}}{z-2011}=-\frac{3}{4}\)

\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)\)

\(\left(\frac{1}{z-2011}-\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^2=0\\\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2=0\\\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-2009}}=\frac{1}{2}\\\frac{1}{\sqrt{y-2010}}=\frac{1}{2}\\\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)( tmđk )

b. ĐK : x² - 9 \(\ge\)0 <=> x²\(\ge\)9 <=> - 3\(\le\)x\(\le\)3

\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

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

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-3}=0\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\left(tmdk\right)\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\)

TH :\(\sqrt{x+3}+\sqrt{x-3}=0\)

Vì \(\sqrt{x+3}+\sqrt{x-3}\ge0\forall x\). Dấu "=" xảy ra <=> \(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=0\\\sqrt{x-3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)( mâu thuẫn )

Vậy pt có nghiệm duy nhất là x = 3

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

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

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

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

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

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