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

giúp mik vs cảm ơn mn

a, Với \(x>0;x\ne1\)

 \(P=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)

\(=\left(\frac{x-1}{2\sqrt{x}}\right)^2\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)

\(=\frac{x^2-2x+1}{4x}.\frac{-4\sqrt{x}}{x-1}=\frac{1-x}{\sqrt{x}}\)

Thay x = 4 => \(\sqrt{x}=2\)vào P ta được : 

\(\frac{1-4}{2}=-\frac{3}{2}\)

c, Ta có : \(P< 0\Rightarrow\frac{1-x}{\sqrt{x}}< 0\Rightarrow1-x< 0\)vì \(\sqrt{x}>0\)

\(\Rightarrow-x< -1\Leftrightarrow x>1\)

cho S=1-3+3²-3³+...+3⁹⁸-3⁹⁹                                                                                                                                       

a. Chứng minh: S chia hêt cho 20

b. Rút gọn S, từ đó suy ra 3¹⁰⁰ chia 4 dư 1

chịu

a. ĐKXĐ: 

\(\hept{\begin{cases}\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}}\)

b. ta có \(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

c. khi \(x=\frac{1}{4}\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow A=\frac{\frac{1}{2}+1}{\frac{1}{2}}=3\)

khi \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{2}+1\Rightarrow A=\frac{\sqrt{2}+1+1}{\sqrt{2}+1}=\sqrt{2}\)

\(a,ĐKXĐ:A=x\ge0;x\ne1\)

\(b,A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(A=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(A=\frac{\sqrt{x}+1}{\sqrt{x}}< =>ĐPCM\)

c,thay \(x=\frac{1}{4}\)vào A

\(c,A=\frac{\sqrt{\frac{1}{4}}+1}{\sqrt{\frac{1}{4}}}\)

\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}}\)

\(A=3\)

\(x=3+2\sqrt{2}\)

\(x=\sqrt{2}^2+2\sqrt{2}+1\)

\(x=\left(\sqrt{2}+1\right)^2\)thay x vào A

\(A=\frac{\sqrt{\left(\sqrt{2}+1\right)^2}+1}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(A=\frac{\sqrt{2}+1+1}{\sqrt{2}+1}\)

\(A=\frac{2+\sqrt{2}}{\sqrt{2}+1}\)

\(A=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\)