Question

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

<=> ( 5x - 10\(\sqrt{x}\) - \(\sqrt{x}\)+ 2 ) - ( 4x - \(4\sqrt{x}\)+ 1 ) = 1

<=> 5x - 10\(\sqrt{x}\) - \(\sqrt{x}\)+ 2 - 4x + \(4\sqrt{x}\)+ 1 -1 = 0

<=> x -7\(\sqrt{x}\) + 2= 0

xong rồi làm thế nào nữa ạ ?

Answer 1

Tacó \(\Delta\)=(-7)²-4x1x2=41>0 =>\(\sqrt{_{ }x1}\)=\(\dfrac{7+\sqrt{41}}{2}\)=>\(_{x1}\)=\(\dfrac{\left(7+\sqrt{41}\right)^2}{4}\)=\(\dfrac{45+7\sqrt{41}}{2}\) =>\(\sqrt{_{ }x2}\)=\(\dfrac{7-\sqrt{41}}{2}\)=>\(_{x_2}\)=\(\dfrac{\left(7-\sqrt{41^{ }}\right)^2}{4}\)=\(\dfrac{45-7\sqrt{41}}{2}\) so sánh với điều kiện X>_0

Answer 2

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

<=> ( 5x- 10\(\sqrt{x}\) - \(\sqrt{x}\) + 2 ) - ( 4x - \(4\sqrt{x}\) + 1 ) = 1

<=> 5x- 10\(\sqrt{x}\) - \(\sqrt{x}\) + 2 - 4x + \(4\sqrt{x}\) - 1- 1 = 0

<=> x - \(7\sqrt{x}\) = 0

<=> \(\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-7=0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}x=0\\\sqrt{x=7}\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}x=0\\x=49\end{matrix}\right.\)

Vậy phương trình có nghiệm là 0 và 49