Câu hỏi của Lâm Tinh Thần

Question

$\sqrt{x-1}+\sqrt{2x-1}=5$ giải pt giúp mình nha

Ann · Accepted Answer

$\sqrt{x-1}+\sqrt{2x-1}=5$

$\Leftrightarrow\left(\sqrt{x-1}-2\right)+\left(\sqrt{2x-1}-3\right)=0$

$\Leftrightarrow\dfrac{x-1-4}{\sqrt{x-1}+2}+\dfrac{2x-1-9}{\sqrt{2x-1}+3}=0$

$\Leftrightarrow\dfrac{x-5}{\sqrt{x-1}+2}+\dfrac{2\left(x-5\right)}{\sqrt{2x-1}+3}=0$

$\Leftrightarrow\left(\dfrac{1}{\sqrt{x-1}+2}+\dfrac{2}{\sqrt{2x-1}+3}\right)\left(x-5\right)=0$

$\Rightarrow x=5$Câu trả lời: $\sqrt{x-1}+\sqrt{2x-1}=5$

$\Leftrightarrow\left(\sqrt{x-1}-2\right)+\left(\sqrt{2x-1}-3\right)=0$

$\Leftrightarrow\dfrac{x-1-4}{\sqrt{x-1}+2}+\dfrac{2x-1-9}{\sqrt{2x-1}+3}=0$

$\Leftrightarrow\dfrac{x-5}{\sqrt{x-1}+2}+\dfrac{2\left(x-5\right)}{\sqrt{2x-1}+3}=0$

$\Leftrightarrow\left(\dfrac{1}{\sqrt{x-1}+2}+\dfrac{2}{\sqrt{2x-1}+3}\right)\left(x-5\right)=0$

$\Rightarrow x=5$

Chương I - Căn bậc hai. Căn bậc ba