`\sqrt(9x^2-6x+1)=5`
`⇔(3x-1)^2=25`
`⇔`\(\left[ \begin{array}{l}3x-1=5\\3x-1=-5\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=-4/3\end{array} \right.\)
`2`
`5\sqrt(3-x)-0,5.\sqrt(12-4x)=8`
`ĐK:x≤3`
`⇔5\sqrt(3-x)-0,5.2.\sqrt(3-x)=8`
`⇔4\sqrt(3-x)=8`
`⇔\sqrt(3-x)=2`
`⇔3-x=4`
`⇔x=-1`
`3`
`\sqrt(x+4-4√x)+2√x=7`
`⇔\sqrt((√x-2)^2)=7-2√x`
`⇔`\(\left[ \begin{array}{l}√x-2=7-2√x\\√x-2=2√x-7\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}√x=3\\√x=5\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=9\\x=25\end{array} \right.\)
`4`
`\sqrt(x-3)+\sqrt(y-5)=10-4/(\sqrt(x-3))-9/(\sqrt(y-5))`
`⇔\sqrt(x-3)+\sqrt(y-5)+4/(\sqrt(x-3))+9/(\sqrt(y-5))=10`
ta có
`\sqrt(x-3)+4/(\sqrt(x-3))+\sqrt(y-5)+9/(\sqrt(y-5))≥2\sqrt(\sqrt(x-3) 4/(\sqrt(x-3)))+2\sqrt(\sqrt(y-5) 9/(\sqrt(y-5)) )≥2.2+3.3≥10`
`''=''`xẩy ra khi
`x-3=1` và `y-5=3`
`⇔x=4 `và `y=8`
mà `\sqrt(x-3)+\sqrt(y-5)+4/(\sqrt(x-3))+9/(\sqrt(y-5))=10`
`⇒x=4 `và `y=8`
