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

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1. Đáp án:

    $x=6$

   Giải thích các bước giải:

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

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

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

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

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

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

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

   $⇔x=6(T/M)$

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