Question

Ai đó giải giúp e vơi

Answer 1

\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow4\sqrt{2x-1}+3\sqrt{2x-1}=4\\ \Leftrightarrow\sqrt{2x-1}=\dfrac{4}{7}\\ \Leftrightarrow2x-1=\dfrac{16}{49}\\ \Leftrightarrow x=\dfrac{65}{98}\left(tm\right)\)

Answer 2

\(\sqrt{32x-16}+\sqrt{18x-9}=4\) (ĐKXĐ: x≥\(\dfrac{1}{2}\))

⇔ \(\sqrt{16\left(2x-1\right)}+\sqrt{9\left(2x-1\right)}=4\)

⇔ 4\(\sqrt{2x-1}\)\(+3\sqrt{2x-1}\)= 4

⇔ 7\(\sqrt{2x-1}=4\)

⇔ \(\sqrt{2x-1}=\dfrac{4}{7}\)

⇔ \(2x-1=\dfrac{16}{49}\)

⇔ 2x = \(\dfrac{65}{49}\)

⇔ x = \(\dfrac{65}{98}\) (TM)

Vậy x = \(\dfrac{65}{98}\)