Question

Câu 2. (1,5 điểm) Giải các phương trình sau:

a) $\dfrac{5}{3} \sqrt{9 x+18}+\dfrac{1}{2} \sqrt{4 x+8}-15=\sqrt{2+x}$.

b) $\sqrt{x^{2}-4 x+4}-6=2 x$.

Answer 1

Answer 2

Answer 3

Bài 2:

a) \(\dfrac{5}{3}\sqrt{9x+18}+\dfrac{1}{2}\sqrt{4x+8}-15=\sqrt{2+x}\) (ĐKXĐ: x ≥ -2)

⇔ \(5\sqrt{x+2}+\sqrt{x+2}-\sqrt{x+2}=15\)

⇔ \(5\sqrt{x+2}=15\)

⇔ \(\sqrt{x+2}=3\)

⇔ \(x+2=9\)

⇔ \(x=7\left(TM\right)\)

Vậy S = {7}

b) \(\sqrt{x^2-4x+4}-6=2x\) (ĐKXĐ: x ≥ -3)

⇔ \(\sqrt{\left(x-2\right)^2}=2x+6\)

⇔ \(\left|x-2\right|=2x+6\)

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

⇔ ​\(\left\{{}\begin{matrix}x-2-2x-6=0\\x-2+2x+6=0\end{matrix}\right.\)​

⇔\(\left\{{}\begin{matrix}-x-8=0\\3x+4=0\end{matrix}\right.\)  ⇔ \(\left\{{}\begin{matrix}x=-8\left(KTM\right)\\x=-\dfrac{4}{3}\left(TM\right)\end{matrix}\right.\) 

Vậy S = \(\left\{-\dfrac{4}{3}\right\}\)