Question

Giải phương trình:

a) \(x^2+x=36-12\sqrt{x+1}\)

b) \(4\sqrt{x+1}=x^2-5x+14\)

c) \(\dfrac{\left(\sqrt{x}-1\right)^2}{2}=2\sqrt{x}-4-\sqrt{x-9}\)

Answer 1

Câu b : \(x^2-5x+14=4\sqrt{x+1}\) ( ĐK : \(x\ge-1\) )

\(\Leftrightarrow x^2-5x+14-4\sqrt{x+1}=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left[\left(x+1\right)-4\sqrt{x+1}+4\right]=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)

Do : \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(\sqrt{x+1}-2\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(\sqrt{x+1}-2\right)^2=0\end{matrix}\right.\Leftrightarrow x=3\)

Vậy \(x=3\)

Answer 2

a. Ta có : x² + x = 36 - 12\(\sqrt{x+1}\)

⇌ x² + 2x + 1 = 36 - 12\(\sqrt{x+1}\) + x + 1

⇌ (x+1)² = ( \(\sqrt{x+1}\) -6)²

⇌ (x+1)² - ( \(\sqrt{x+1}\) -6)² = 0

còn lại tự làm nha