b)\(\sqrt{x^2-10x+25}=2x-3\) ĐK:x≥3/2
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=2x-3\)
\(\Leftrightarrow\left|x-5\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x-3\\x-5=3-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy phương trình có 2 nghiệm là ...
a: Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}-4\sqrt{x+3}=2\)
\(\Leftrightarrow x+3=4\)
hay x=1
c)ĐK:x≥1/2
\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-\sqrt{2x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)
\(\Leftrightarrow\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|=2\) (do \(\sqrt{2x-1}+1\ge0\))
\(\Leftrightarrow\sqrt{2x-1}+1+\left|1-\sqrt{2x-1}\right|=2\)
Ta có:\(\sqrt{2x-1}+1+\left|1-\sqrt{2x-1}\right|\ge\sqrt{2x-1}+1+1-\sqrt{2x-1}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\)
\(\Leftrightarrow\dfrac{1}{2}\le x\le1\)
Vậy phương trình có tâp nghiệm \(S=\left\{x|\text{}\dfrac{1}{2}\le x\le1\right\}\)
