`3b)\sqrt{25x-25}-15/2\sqrt{(x-2)/9}=6+3/2\sqrt{x-1}`
ĐK:`x>=1`
`pt<=>sqrt{25(x-1)}-15/2*1/3sqrt{x-1}-3/2sqrt{x-1}=6`
`<=>5sqrt{x-1}-5/2sqrt{x-1}-3/2sqrt{x-1}=6`
`<=>5sqrt{x-1}-4sqrt{x-1}=6`
`<=>sqrt{x-1}=6`
`<=>x-1=36`
`<=>x=37(tmddk)`
Vậy `S={37}`
3b) \(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\dfrac{3}{2}\sqrt{x-1}\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt{25\left(x-1\right)}-\dfrac{15}{2}\sqrt{\dfrac{1}{9}.\left(x-1\right)}-\dfrac{3}{2}\sqrt{x-1}=6\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}.\dfrac{1}{3}\sqrt{x-1}-\dfrac{3}{2}\sqrt{x-1}=6\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{x-1}=6\)
\(\Leftrightarrow\sqrt{x-1}=6\Rightarrow x-1=36\Rightarrow x=37\)
Bài 3:
c) ĐKXĐ: \(x\ge0\)
Ta có: \(\sqrt{4+5\sqrt{x}}=3\)
\(\Leftrightarrow5\sqrt{x}+4=9\)
\(\Leftrightarrow5\sqrt{x}=5\)
\(\Leftrightarrow x=1\)
Vậy: S={1}
d) Ta có: \(x-\sqrt{4x-20}=20\)
\(\Leftrightarrow\sqrt{4x-20}=x-20\)
\(\Leftrightarrow\left(x-20\right)^2=4x-20\)
\(\Leftrightarrow x^2-40x+400-4x+20=0\)
\(\Leftrightarrow x^2-44x+420=0\)
\(\Delta=\left(-44\right)^2-4\cdot1\cdot420=256\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{44-16}{2}=\dfrac{28}{2}=14\\x_2=\dfrac{44+16}{2}=\dfrac{60}{2}=30\end{matrix}\right.\)
Vậy: S={14;30}
