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Tìm x, biết:

$\sqrt{25x-25}-\dfrac{15}{2}.\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}$

qwerty · Accepted Answer

$\sqrt{25x-25}-\dfrac{15}{2}\cdot\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}$ (1)

$\Leftrightarrow\sqrt{25\left(x-1\right)}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}$

$\Leftrightarrow\sqrt{25}\sqrt{x-1}-\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}$

$\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}$

$\Leftrightarrow\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}$

$\Leftrightarrow5\sqrt{x-1}=12+2\sqrt{x-1}$

$\Leftrightarrow5\sqrt{x-1}-2\sqrt{x-1}=12$

$\Leftrightarrow3\sqrt{x-1}=12$

$\Leftrightarrow\sqrt{x-1}=4$

$\Leftrightarrow x-1=16$

$\Leftrightarrow x=16+1$

$\Leftrightarrow x=17$

Vậy tập nghiệm phương trình (1) là $S=\left\{17\right\}$Câu trả lời: $\sqrt{25x-25}-\dfrac{15}{2}\cdot\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}$ (1)