4, $\sqrt[]{2x^{2}+5x-3}-\sqrt[]{2x-1}=0$ ĐKXĐ: $x≥\frac{1}{2}$
⇔$\sqrt[]{(2x-1)(x+3)} - \sqrt[]{2x-1}=0$
⇔$\sqrt[]{2x-1}(\sqrt[]{x+3}-1)=0$
⇔$\left \{ {{\sqrt[]{2x-1}=0} \atop {\sqrt[]{x+3}=1}} \right.$
⇔$\left \{ {{2x-1=0} \atop {x+3=1}} \right.$
⇔$\left \{ {{x=\frac{1}{2}(t/m)} \atop {x=-2}(Loại)} \right.$
⇔$x=\frac{1}{2}$
5,$\sqrt[]{x^{2}-9}-\sqrt[]{x^{2}-6x+9}=0$ ĐKXĐ: \(\left[ \begin{array}{l}x≤-3\\x≥3\end{array} \right.\)
⇔$\sqrt[]{(x+3)(x-3)}-\sqrt[]{(x-3)^{2}}=0$
⇔$\sqrt[]{x-3}(\sqrt[]{x+3}-\sqrt[]{x-3})=0$
⇔$\left \{ {{\sqrt[]{x-3}=0} \atop {\sqrt[]{x+3}=\sqrt[]{x-3}}} \right.$
⇔$\left \{ {{x-3=0} \atop {x+3=x-3}} \right.$
⇔$\left \{ {{x=3(t/m)} \atop {x=-2}(Loại)} \right.$
⇔$x=3$
6, $\sqrt[]{x^{2}-5x+6}+\sqrt[]{x+1}=\sqrt[]{x-2}+\sqrt[]{x^{2}-2x-3}$ ĐKXĐ: $x≥3$
⇔$\sqrt[]{x^{2}-5x+6}-\sqrt[]{x-2}=\sqrt[]{x^{2}-2x-3}-\sqrt[]{x+1}$
⇔$\sqrt[]{(x-2)(x-3)}-\sqrt[]{x-2}=\sqrt[]{(x+1)(x-3)}-\sqrt[]{x+1}$
⇔$\sqrt[]{x-2}(\sqrt[]{x-3}-1)=\sqrt[]{x+1}(\sqrt[]{x-3}-1)$
⇔$(\sqrt[]{x-3}-1)(\sqrt[]{x+1}-\sqrt[]{x-2})=0$
⇔$\left \{ {{\sqrt[]{x-3}-1=0} \atop {\sqrt[]{x+1}-\sqrt[]{x-2}=0}} \right.$
⇔$\left \{ {{\sqrt[]{x-3}=1} \atop {\sqrt[]{x+1}=\sqrt[]{x-2}}} \right.$
⇔$\left \{ {{x-3=1} \atop {x+1=x-2}} \right.$
⇔$\left \{ {{x=4}(t/m) \atop {x=\frac{-3}{2}}(loại)} \right.$
⇔$x=4$
