[Lekkoksung]

ข้อ 3
$5(\sqrt{1-x}+\sqrt{1+x} )=6x+8\sqrt{1-x^2}$
แนวคิด
$$5(\sqrt{1-x}+\sqrt{1+x} )=6x+8\sqrt{1-x^2}$$
$$25(2+2\sqrt{1-x^{2}})=36x^{2}+96x\sqrt{1-x^{2}}+64(1-x^{2})$$
$$50+50\sqrt{1-x^{2}}=36x^{2}+96x\sqrt{1-x^{2}}+64-64x^{2}$$
$$28x^{2}=96x\sqrt{1-x^{2}}-50\sqrt{1-x^{2}}+14$$
ให้ $u=\sqrt{1-x^{2}}$ แล้วยังไงต่อครับ