Question

[....]

giải các pt sau

x\(^2+5x+\sqrt{x^2+5x+4}=2\)

\(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\)

Answer 1

[Edogawa Conan]

a,ĐK: x≥-1

Đặt \(t=\sqrt{x^2+5x+4}\left(t\ge0\right)\)

  ⇒ \(t^2+t-6=0\)

  \(\Leftrightarrow\left(t+3\right)\left(t-2\right)=0\)

  \(\Leftrightarrow\left[{}\begin{matrix}t=-3\left(loại\right)\\t=2\end{matrix}\right.\)

  \(\Leftrightarrow\sqrt{x^2+5x+4}=2\)

  \(\Leftrightarrow x^2+5x+4=4\)

  \(\Leftrightarrow x\left(x+5\right)=0\)

  \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-5\left(loại\right)\end{matrix}\right.\)

Answer 2

[Edogawa Conan]

b,ĐK: \(0\le x\le2\)

Ta có: \(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\)

    \(\Leftrightarrow-x^2-3x+10=3\sqrt{x^2+3x}\)     (1)

Đặt \(t=\sqrt{x^2+3x}\left(t\ge0\right)\)

  \(\Rightarrow\left(1\right)\Leftrightarrow-t^2+10-3t=0\)  

             \(\Leftrightarrow\left(t+5\right)\left(2-t\right)=0\)

             \(\Leftrightarrow\left[{}\begin{matrix}t=-5\left(loại\right)\\t=2\end{matrix}\right.\)

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

             \(\Leftrightarrow x^2+3x=4\)

             \(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

             \(\Leftrightarrow\left[{}\begin{matrix}x=-4\left(loại\right)\\x=1\left(tm\right)\end{matrix}\right.\)