\(\sqrt{1+x}=\sqrt{6-x}-\sqrt{5-2x}\)

\(\Leftrightarrow\sqrt{1+x}+\sqrt{5-2x}=\sqrt{6-x}\)

\(\Rightarrow\left(\sqrt{1+x}+\sqrt{5-2x}\right)^2=\left(\sqrt{6-x}\right)^2\)

\(\Leftrightarrow1+x+5-2x+2\sqrt{\left(1+x\right)\left(5-2x\right)}=6-x\)

\(\Leftrightarrow6-x+2\sqrt{\left(1+x\right)\left(5-2x\right)}=6-x\)

\(\Leftrightarrow2\sqrt{\left(1+x\right)\left(5-2x\right)}=0\)

\(\Leftrightarrow\orbr{\begin{cases}1+x=0\\5-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{2}\end{cases}}}\)

Vậy ..

thêm dòng đầu là \(ĐKXĐ:-1\le x\le\frac{5}{2}\)

pt <=> \(\hept{\begin{cases}x+1\ge0\\1+x\sqrt{x^2+4}=x^2+2x+1\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\left(1\right)\\x\sqrt{x^2+4}=x^2+2x\left(2\right)\end{cases}}\)

Ta có: (2) <=> \(\orbr{\begin{cases}x=0\left(tm\left(1\right)\right)\\\sqrt{x^2+4}=x+2\end{cases}}\)

Với \(\sqrt{x^2+4}=x+2\Leftrightarrow\hept{\begin{cases}x^2+4=x^2+4x+4\\x\ge-2\end{cases}}\)<=> x = 0 ( tm (1))

Vậy x = 0

Nhận xét : \(\sqrt{\left(5-2\sqrt{6}\right)^x}.\sqrt{\left(5+2\sqrt{6}\right)^x}=1\)

Ta đặt \(\sqrt{\left(5-2\sqrt{6}\right)^x}=a\Rightarrow\sqrt{\left(5+2\sqrt{6}\right)^x}=\frac{1}{a}\)

Khi đó phương trình ban đầu trở thành :

\(a+\frac{1}{a}=10\Rightarrow a^2-10a+1=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=5+2\sqrt{6}\\a=5-2\sqrt{6}\end{cases}}\)

+) Với \(a=5+2\sqrt{6}\Rightarrow\sqrt{\left(5-2\sqrt{6}\right)^x}=5+2\sqrt{6}\)

\(\Leftrightarrow\left(5-2\sqrt{6}\right)^x=\left(5+2\sqrt{6}\right)^2=\left(\frac{1}{5-2\sqrt{6}}\right)^2\)

\(\Leftrightarrow x=-2\)

+) Với \(a=5-2\sqrt{6}\Rightarrow\sqrt{\left(5-2\sqrt{6}\right)^x}=5-2\sqrt{6}\)

\(\Leftrightarrow\left(5-2\sqrt{6}\right)^x=\left(5-2\sqrt{6}\right)^2\)

\(\Leftrightarrow x=2\)

Vậy \(x\in\left\{-2,2\right\}\) thỏa mãn đề.

loanwngf ngoằng quá

mk biết

E max 

\(\Leftrightarrow\frac{1}{2x-\sqrt{x}+5}\) lớn nhất 

\(2x-\sqrt{x}+5\)   nhỏ nhất 

\(=\left(\sqrt{2x}\right)^2-2\cdot\sqrt{2x}\cdot\frac{\sqrt{2}}{4}+\left(\frac{\sqrt{2}}{4}\right)^2-\left(\frac{\sqrt{2}}{4}\right)^2+5\) 

\(=\left(\sqrt{2x}-\frac{\sqrt{2}}{4}\right)^2+\frac{39}{8}\) 

Ta có \(\left(\sqrt{2x}-\frac{\sqrt{2}}{4}\right)^2+\frac{39}{8}\ge\frac{39}{8}\forall x\ge0\) 

Dấu = xảy ra 

\(\Leftrightarrow\left(\sqrt{2x}-\frac{\sqrt{2}}{4}\right)^2=0\) 

\(\sqrt{2}\cdot\sqrt{x}-\frac{\sqrt{2}}{4}=0\) 

\(\sqrt{2}\cdot\sqrt{x}=\frac{\sqrt{2}}{4}\) 

\(\sqrt{x}=\frac{\sqrt{2}}{4}:\sqrt{2}\) 

\(\sqrt{x}=\frac{1}{4}\) 

\(x=\left(\frac{1}{4}\right)^2=\frac{1}{16}\) 

E max = \(\frac{1}{\frac{39}{8}}=\frac{8}{39}\Leftrightarrow x=\frac{1}{16}\)

\(E=\frac{1}{2x-\sqrt{x}+5}\)

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

\(=\frac{1}{2\left(x-2.\sqrt{x}.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{5}{2}\right)}\)

\(=\frac{1}{2\left(x-\frac{\sqrt{x}}{4}\right)^2+\frac{39}{8}}\le\frac{8}{39}\)

Dấu "="xảy ra \(\Leftrightarrow x-\frac{\sqrt{x}}{4}=0\Leftrightarrow x=\frac{\sqrt{x}}{4}\)

\(\Leftrightarrow16x^2=x\Leftrightarrow x\left(16x-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{16}\end{cases}}\)

Vậy \(E_{max}=\frac{8}{39}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{16}\end{cases}}\)

\(ĐKXĐ:x\ge-\frac{3}{2}\)

Ta có : \(2\sqrt{x+\sqrt{2x+3}+2}=\sqrt{2}\left(x+1\right)\)

\(\Leftrightarrow2.\sqrt{2}.\sqrt{x+\sqrt{2x+3}+2}=\sqrt{2}.\sqrt{2}.\left(x+1\right)\)

\(\Leftrightarrow2\sqrt{2x+2\sqrt{2x+3}+4}=2\left(x+1\right)\)

\(\Leftrightarrow2\sqrt{\left(2x+3\right)+2\sqrt{2x+3}+1}=2.\left(x+1\right)\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x+3}+1\right)^2}=x+1\)

\(\Leftrightarrow\sqrt{2x+3}+1=x+1\)

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

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\2x+3=x^2\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2-2x-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\\left(x-3\right)\left(x+1\right)=0\end{cases}}\) \(\Leftrightarrow x=3\)( Thỏa mãn )

Vậy pt có nghiệm duy nhất \(x=3\)

mik ko biết

Để y có nghĩa

\(\Leftrightarrow\hept{\begin{cases}x^2-5x+6\ge0\\x-1\ge0\\\sqrt{x-1}\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2-5x+25-19\ge0\\x\ge1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-5\right)^2-19\ge0\\x\ge1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-5\right)^2\ge19\\x\ge1\end{cases}}\)

Đến đây tự làm được rồi nhỉ ??