Question

Giải các pt sau:

1. x-\(\sqrt{2x-5}\)= 4

2. 2x²-3-5\(\sqrt{2x^2+3}\)= 5

3. 2x²+3x+3=5\(\sqrt{2x^2+3x+9}\)

Answer 1

1) x-\(\sqrt{2x-5}\)=4

ĐK: \(\left\{{}\begin{matrix}2x-5\ge0\\x\ge4\end{matrix}\right.\)=> x\(\ge\)4

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

bình phương hai vế:

\(x^2-8x+16\) =2x-5

<=>\(x^2\) -10x+21=0 <=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

2) \(2x^2-3-5\sqrt{2x^2+3}=0\)(*)

ĐK:\(2x^2-3>0\Leftrightarrow x^2>\dfrac{3}{2}\)

<=>\(\left[{}\begin{matrix}x>\sqrt{\dfrac{3}{2}}\\x< -\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)

(*)<=>

Answer 2

cau 2 là bằng 0 ko phải bằng 5 nha