Đáp án:
$a.x=\dfrac{11\pm\sqrt[]{17}}{2}$
$b. x=-1\quad hoặc \quad x=3$
Giải thích các bước giải:
$a.\sqrt[]{-x^2+11x-24}=-x^2+11x-26$
$\rightarrow(-x^2+11x-24)-\sqrt[]{-x^2+11x-24}-2=0$
$\rightarrow(-x^2+11x-24)-1-(\sqrt[]{-x^2+11x-24}+1)=0$
$\rightarrow(\sqrt[]{-x^2+11x-24}-1)(\sqrt[]{-x^2+11x-24}+1)-(\sqrt[]{-x^2+11x-24}+1)=0$
$\rightarrow(\sqrt[]{-x^2+11x-24}-2)(\sqrt[]{-x^2+11x-24}+1)=0$
$\rightarrow \sqrt[]{-x^2+11x-24}-2=0$
$\rightarrow \sqrt[]{-x^2+11x-24}=2$
$\rightarrow -x^2+11x-24=2$
$\rightarrow x^2-11x+26=0$
$\rightarrow x^2-2x\dfrac{11}{2}+\dfrac{11^2}{4}=\dfrac{17}{4}$
$\rightarrow (x-\dfrac{11}{2})^2=\dfrac{17}{4}$
$\rightarrow x-\dfrac{11}{2}=\pm\dfrac{\sqrt[]{17}}{2}$
$\rightarrow x=\dfrac{11\pm\sqrt[]{17}}{2}$
$b.\sqrt[]{3x+7}-\sqrt[]{x+1}=2$
$\rightarrow \sqrt[]{3x+7}=\sqrt[]{x+1}+2$
$\rightarrow 3x+7=(\sqrt[]{x+1}+2)^2$
$\rightarrow 3x+7=x+1+4\sqrt[]{x+1}+4$
$\rightarrow 2x+2=4\sqrt[]{x+1}$
$\rightarrow (x+1)-2\sqrt[]{x+1}=0$
$\rightarrow \sqrt[]{x+1}(\sqrt[]{x+1}-2)=0$
$\rightarrow \sqrt[]{x+1}=0\quad hoặc \quad \sqrt[]{x+1}-2=0$
$\rightarrow x=-1\quad hoặc \quad x=3$
