a ) \(\sqrt{2x-1}=5\left(ĐKXĐ:x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x-1=25\)
\(\Leftrightarrow2x=24\)
\(\Leftrightarrow x=12\left(tmđk\right)\)
Vậy \(x=12\)
b ) \(\sqrt{\dfrac{x^2}{4}+x+1}=2x-1\)
\(\Leftrightarrow\sqrt{\left(\dfrac{x}{2}+1\right)^2}=2x-1\)
\(\Leftrightarrow|\dfrac{x}{2}+1|=2x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}+1=2x-1\\-\dfrac{x}{2}-1=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=0\end{matrix}\right.\)
Vậy x = \(\dfrac{4}{3}\) hoặc x = 0 .
Bạn tự tìm điều kiện xác định nha!
\(a,\sqrt{2x-1}=5\)
\(\Leftrightarrow\left(\sqrt{2x-1}\right)^2=25\)
\(\Leftrightarrow\left|2x-1\right|=25\)
\(\Leftrightarrow2x-1=25\)
\(\Leftrightarrow2x=26\)
\(\Leftrightarrow x=13\) (tm)
\(b,\sqrt{\dfrac{x^2}{4}+x+1}=2x-1\)
\(\Leftrightarrow\sqrt{\left(\dfrac{x}{2}+1\right)^2}=2x-1\)
\(\Leftrightarrow\dfrac{x}{2}+1=2x-1\)
\(\Leftrightarrow\dfrac{x}{2}-2x=-2\)
\(\Leftrightarrow-\dfrac{5}{2}x=-2\)
\(\Leftrightarrow x=\dfrac{5}{4}\)
Bài a :
\(\sqrt{2x-1}=5\) 
ĐKXĐ \(x\ge\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)}^2=5^2\)
\(\Leftrightarrow\left|2x-1\right|=25\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=25\\1-2x=25\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=13\\x=-12\left(loại\right)\end{matrix}\right.\)\(\)
Vậy \(x=13\)
Mình làm lại ý a nha .
\(\sqrt{2x-1}=5\left(ĐKXĐ:x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x-1=25\)
\(\Leftrightarrow2x=26\)
\(\Leftrightarrow x=13\left(tm\right)\)
Vậy phương trình có nghiệm x = 13 .
