Giải thích các bước giải:
a.$\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{6+2x}{\sqrt{5+x}}=\dfrac{8}3$
$\to\dfrac{2(5-x)-4}{\sqrt{5-x}}+\dfrac{2(5+x)-4}{\sqrt{5+x}}=\dfrac{8}3$
$\to 2\sqrt{5-x}-\dfrac{4}{\sqrt{5-x}}+2\sqrt{5+x}-\dfrac{4}{\sqrt{5+x}}=\dfrac{8}3$
$\to \sqrt{5-x}-\dfrac{2}{\sqrt{5-x}}+\sqrt{5+x}-\dfrac{2}{\sqrt{5+x}}=\dfrac{4}3$
$\to (\sqrt{5-x}+\sqrt{5+x})-2(\dfrac{1}{\sqrt{5-x}}+\dfrac{1}{\sqrt{5+x}})=\dfrac{4}3$
Đặt $\sqrt{5-x}=a,\sqrt{5+x}=b\to a^2+b^2=10\to (a+b)^2-2ab=10\to ab=\dfrac{(a+b)^2-10}{2}$
$\to (a+b)-2(\dfrac 1a+\dfrac 1b)=\dfrac{4}3$
$\to (a+b)-2\dfrac{a+b}{ab}=\dfrac{4}3$
$\to (a+b)-2\dfrac{a+b}{\dfrac{(a+b)^2-10}{2}}=\dfrac{4}3$
$\to a+b=4,\dfrac{-4+\sqrt{46}}{3}$ vì $a,b>0$
$\to ab\to $Tìm x
b.Xem lại đề
