\(x-\sqrt{2x-9}=6\left(Đk:x\ge\frac{9}{2}\right)\)

\(\Rightarrow\sqrt{2x-9}=x-6\left(đk:x\ge6\right)\)

\(\Rightarrow\left(\sqrt{2x-9}\right)^2=\left(x-6\right)^2\)

\(\Rightarrow2x-9=x^2-12x+36\)

\(\Rightarrow-x^2+14x-45=0\)

\(\Rightarrow-x^2+9x+5x-45=0\)

\(\Rightarrow-x\left(x-9\right)+5\left(x-9\right)=0\)

\(\Rightarrow\left(x-9\right)\left(5-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=9\\x=5\end{cases}\left(TM\right)}\)

\(a,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)

\(ĐK:x\ne\pm1\)

\(PT\Leftrightarrow\left(2x+1\right)\left(x-1\right)=5\left(x-1\right)^2\)

\(\Leftrightarrow2x^2-x-1=5\left(x^2-2x+1\right)\)

\(\Leftrightarrow3x^2-9x+4=0\)

\(\Leftrightarrow3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)=\frac{11}{4}\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{11}{12}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{11}{12}}+\frac{3}{2}\\x=-\sqrt{\frac{11}{12}}+\frac{3}{2}\end{cases}}\)

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

mn ơi giúp e

e sẽ k cho ai trả lời đúng

a) Xét 2 tam giác vuông DHC và FBC có: ^HCD chung => \(\Delta DHC~\Delta FBC\)

=> \(\frac{CD}{CF}=\frac{CH}{BC}\) => \(CH.CF=BC.CD\) (1) 

tương tự với 2 tam giác vuông DBH và EBC có: ^EBC chung => \(\Delta DBH~\Delta EBC\)

=> \(\frac{BD}{BE}=\frac{BH}{BC}\) => \(BH.BE=BC.BD\) (2) 

(1) và (2) => \(CH.CF+BH.BE=BC\left(BD+CD\right)=BC^2\)

b) CM tương tự câu a), ta cũng có: \(AH.AD+BH.BE=AB^2;AH.AD+CH.CF=AC^2\)

cộng lại ta có đpcm