\(\sqrt{X}\)= -a

Mình mới học lớp 8 lên làm sợ sai ý

\(\)<=> \(\left(x-5\right)\left(x+4-\frac{64}{18+2\sqrt{1+16}}\right)=0\)  

      <=>\(x-5=0\) 

        <=>\(x=5\)

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

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

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

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

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

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

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

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.