giải PT sau : $\sqrt[]{x-5}+ \sqrt[4]{7-x}=2$

Đáp án:

 $x=6$

Giải thích các bước giải:

 $\sqrt[]{x-5}+\sqrt[4]{7-x}=2$  $ĐKXĐ: 5≤x≤7$

$⇔(\sqrt[]{x-5}-1)+(\sqrt[4]{7-x}-1)=0$

$⇔\dfrac{(\sqrt[]{x-5}-1).(\sqrt[]{x-5}+1)}{\sqrt[]{x-5}+1}+\frac{(\sqrt[4]{7-x}-1).(\sqrt[4]{7-x}+1).(\sqrt[]{7-x}+1)}{(\sqrt[4]{7-x}+1).(\sqrt[]{7-x}+1)}=0$

$⇔\dfrac{x-5-1}{\sqrt[]{x-5}+1}+\frac{(\sqrt[]{7-x}-1)(\sqrt[]{7-x}+1)}{(\sqrt[4]{7-x}+1).(\sqrt[]{7-x}+1)}=0$

$⇔\dfrac{x-6}{\sqrt[]{x-5}+1}+\frac{7-x-1}{(\sqrt[4]{7-x}+1).(\sqrt[]{7-x}+1)}=0$

$⇔\dfrac{x-6}{\sqrt[]{x-5}+1}+\frac{6-x}{(\sqrt[4]{7-x}+1).(\sqrt[]{7-x}+1)}=0$

$⇔\begin{cases}x-6=0\\6-x=0\end{cases}$

$⇔x=6(T/M)$