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

Aj giải giúp tui với.....! :-(

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

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

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

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

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

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

  VẬY       \(x=2013;y=2014;z=2015\)

ko biet E=MC'2

tui cũng ko biết làm nữa

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

Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)

Ta có: \(\frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0\)

\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0\)

\(\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow a=b=c=\frac{1}{2}\)

Thay vào tìm x;y;z

Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)

Ta có: \frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0a1​−a21​+b1​−b21​+c1​−c21​−43​=0

\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0⇔a21​−a1​+b21​−b1​+c21​−c1​+43​=0

\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0⇔(a21​−a1​+41​)+(b21​−b1​+41​)+(c21​−c1​+41​)=0

\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0⇔(a1​−21​)2+(b1​−21​)2+(c1​−21​)2=0

\Leftrightarrow a=b=c=\frac{1}{2}⇔a=b=c=21​

Thay vào tìm x;y;z

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

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