\(\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{60}{7-x}}=6\)
\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-\sqrt{\dfrac{126}{14}}+\sqrt{\dfrac{60}{7-x}}-\sqrt{\dfrac{45}{5}}=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-\dfrac{126}{14}}{\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{126}{14}}}+\dfrac{\dfrac{60}{7-x}-\dfrac{45}{5}}{\sqrt{\dfrac{60}{7-x}}+\sqrt{\dfrac{45}{5}}}=0\)
\(\Leftrightarrow\dfrac{\dfrac{-3\left(3x-1\right)}{x-5}}{\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{126}{14}}}+\dfrac{\dfrac{-3\left(3x-1\right)}{x-7}}{\sqrt{\dfrac{60}{7-x}}+\sqrt{\dfrac{45}{5}}}=0\)
\(\Leftrightarrow-3\left(3x-1\right)\left(\dfrac{\dfrac{1}{x-5}}{\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{126}{14}}}+\dfrac{\dfrac{1}{x-7}}{\sqrt{\dfrac{60}{7-x}}+\sqrt{\dfrac{45}{5}}}\right)=0\)
Dễ thấy: \(\dfrac{\dfrac{1}{x-5}}{\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{126}{14}}}+\dfrac{\dfrac{1}{x-7}}{\sqrt{\dfrac{60}{7-x}}+\sqrt{\dfrac{45}{5}}}>0\)
\(\Rightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
gì mà kiểu khủng bố thê nhỉ
(rất may x =1/3 là nghiệm)
\(\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{60}{7-x}}=6\) (1)
đk: \(\left\{{}\begin{matrix}\dfrac{42}{5-x}\ge0\\\dfrac{60}{7-x}\ge0\end{matrix}\right.\) \(\Rightarrow x< 5\)
\(\left(1\right)\Leftrightarrow\left[\sqrt{\dfrac{42}{5-x}}-3\right]+\left[\sqrt{\dfrac{60}{7-x}}-3\right]=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{42}{7-x}}+3}=0\)
\(\Leftrightarrow\dfrac{-3+9x}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{-3+9x}{\left(7-x\right)\left(\sqrt{\dfrac{42}{7-x}}+3\right)}=0\)-3+9x =0 => x =1/3 thỏa mãn
x khác 1/3 <=>
\(\Leftrightarrow\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{42}{7-x}}+3\right)}=0\left(2\right)\\\)với đk x< 5 (2) vô nghiệm
kết luận x =1/3 là duy nhất
