Question

[Thầy Tùng Dương]

Giải các phương trình

a) \(\sqrt{x-1-2\sqrt{x-2}}=1;\)

b) \(\sqrt{x+\sqrt{x+5}}+\sqrt{x-\sqrt{x+5}}=2\sqrt{2}.\)

Answer 1

[Ĵ❍ƙĔŔᵛᶰ(*•.¸♡ţęąɱ ƒŗęę ƒįŗę❤☆)]

\(a,\sqrt{x-1-2\sqrt{x-2}}=1\)

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}-1=1\\\sqrt{x-2}-1=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=2\\\sqrt{x-2}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x-2}\right)^2=2^2\\\left(\sqrt{x-2}\right)=0^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-2=4\\x-2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=6\\x=2\end{cases}}\)

Answer 2

[Nguyễn Trà My]

a) \(\sqrt{x-1-2\sqrt{x-2}}\)=1

⇔\(\sqrt{x-2-2\sqrt{x-2}+1}\)=1

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

⇔(\(\sqrt{\left(\sqrt{x-2}-1\right)^2}\))²=1²

⇔(\(\sqrt{x-2}\)-1)²=1

⇔\(\left\{{}\begin{matrix}\sqrt{x-2}-1=1\\\sqrt{x-2}-1=-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}\sqrt{x-2}=2\\\sqrt{x-2}=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x-2=4\\x-2=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

      Vậy phương trình có 2 nghiệm là x=6; x=2

b) \(\sqrt{x+\sqrt{x+5}}\)+\(\sqrt{x-\sqrt{x+5}}\)=2\(\sqrt{2}\)    ( đk: x≥-5)

⇔ x+\(\sqrt{x^2-x-5}\)=4

⇔\(\sqrt{x^2-x-5}\)=4-x  

⇔(\(\sqrt{x^2-x-5}\))²= ( 4-x)²

⇔x²-x-5= 16-8x+x²

⇔x²-x+8x-x²=16+5

⇔ 7x=21

⇔x=3 ( thỏa mãn điều kiện xác định)