\(\sqrt{25t^2-9}=2\sqrt{5t-3}\left(t\ge\dfrac{3}{5}\right)\)hoặc\(t\le-\dfrac{3}{5}\))
\(=\sqrt{\left(5t-3\right)\left(5t+3\right)}-2\sqrt{5t-3}=0\)
\(< =>\sqrt{5t-3}\left(\sqrt{5t+3}-2\right)=0\)
\(=>\left[{}\begin{matrix}\sqrt{5t-3}=0\\\sqrt{5t+3}-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}t=0,6\left(TM\right)\\t=0,2\left(loai\right)\end{matrix}\right.\)
vậy t=0,6
\(\sqrt{-2x^2+6}=x-1\)(\(-\sqrt{3}\le x\le\sqrt{3}\) \(\))
\(=>-2x^2+6=x^2-2x+1\)
\(< =>-3x^2+2x+5=0\)
\(\Delta=\left(2\right)^2-4.5.\left(-3\right)=64>0\)
\(=>\left[{}\begin{matrix}x1=\dfrac{-2+\sqrt{64}}{2\left(-3\right)}=-1\left(loai\right)\\x2=\dfrac{-2-\sqrt{64}}{2\left(-3\right)}=\dfrac{5}{3}\left(TM\right)\end{matrix}\right.\)vậy x=5/3
