Giải phương trình sau đây

Question

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

Answer 1

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

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

<=> \({2x-1}=25-10\sqrt{x-1}+x-1\)

\(<=> {2x}=25-10\sqrt{x-1}+x\)

\(<=> 10\sqrt{x-1}=25+x-2x\)

\(<=> 10\sqrt{x-1}=25-x\)

\(<=> 100(x-1)=625-50x+x^2\)

\(<=> 100x-100=625-50x+x^2\)

\(<=> 100x-100-625+50x-x^2=0\)

\(<=> 150x-725-x^2=0\)

\(<=>-x^2+150x-725=0\)

\(<=>x^2-150x+725=0\) (bấm máy)

\(<=>x=5;x=145\)

\(<=>\sqrt{2.5-1}+\sqrt{5-1}=5\)

          \(\sqrt{2.145-1}+\sqrt{145-1}=5\)

\(<=>5=5\) (nhận)

          \(29=5\) (vô lý)

Vậy phương trình này có một nghiệm

 

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Answer 2

\(\sqrt{2x-1}+\sqrt{x-1}=5 (x \geq 1)\\\Leftrightarrow\sqrt{2x-1}-3+\sqrt{x-1}-2=0\\\Leftrightarrow\dfrac{(\sqrt{2x-1})^2-3^2}{\sqrt{2x-1}+3}+\dfrac{(\sqrt{x-1})^2-2^2}{\sqrt{x-1}+2}=0\\\Leftrightarrow\dfrac{2x-10}{\sqrt{2x-1}+3}+\dfrac{x-5}{\sqrt{x-1}+2}=0\\\Leftrightarrow(x-5) \bigg(\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}\bigg)=0\\ Vì \bigg(\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}\bigg)>0\\\Rightarrow x-5=0 \Leftrightarrow x=5 (tm)\)